Question 515 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos4\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos4\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^22\text{x}}{\text{x}^2}$
$=2\lim\limits_{\text{x} \rightarrow0}\Big(\frac{\sin2\text{x}}{\text{x}}\Big)^2$
$=2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin2\text{x}}{2\text{x}}\Big)^2\times(2)^2$
$=2\times1\times4$
$=8$
View full question & answer→Question 525 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sqrt{2\sin\text{x}}-1}{\big(\frac{\pi}{2}-\text{x}\big)^2}$
$\Rightarrow\ \text{x}\rightarrow\frac{\pi}{2},$ then $\frac{\pi}{2}-\text{x}\rightarrow0,$ let $\frac{\pi}{2}-\text{x}=\text{y}$
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}\rightarrow0}=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\sin\big(\frac{\pi}{2}-\text{y}\big)}-1}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)}{\text{y}^2}\times\frac{\big(\sqrt{2-\cos\text{y}}+1\big)}{\big(\sqrt{2-\cos\text{y}}+1\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big({2-\cos\text{y}}-1\big)}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{{1-\cos\text{y}}}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$
$=2\times\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)\times\frac14\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$
$=2\times1\times\frac14\times\frac{1}{1+1}=\frac14$
View full question & answer→Question 535 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}^2+\text{x}^2}-\text{a}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}^2+\text{x}^2}-\text{a}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{\text{a}^2+\text{x}^2}-\text{a}\big)}{\text{x}^2}\times\frac{\big(\sqrt{\text{a}^2+\text{x}^2}+\text{a}\big)}{\big(\sqrt{\text{a}^2+\text{x}^2}+\text{a}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{a}^2+\text{x}^2-\text{a}^2\big)}{\text{x}^2\sqrt{\text{a}^2+\text{x}^2}+\text{a}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{\text{x}^2\big(\sqrt{\text{a}^2+\text{x}^2}+\text{a}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\sqrt{\text{a}^2+\text{x}^2+\text{a}}}$
$=\frac{1}{\text{a}+\text{a}}$
$=\frac{1}{2\text{a}}$
View full question & answer→Question 545 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a}+\text{x})+\sin(\text{a}-\text{x})-2\sin\text{a}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{a}+\text{x}+\text{a}-\text{x}}{2}\big)\cos\big(\frac{\text{a}+\text{x}-\text{a}+\text{x}}{2}\big)-\sin\text{a}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}(\cos\text{x}-1)}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin\text{a}(1-\cos\text{x})}{\text{x}\sin\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{a}\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}\big)\text{x}}$
$=-2\sin\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{\tan\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\times\frac12$
$=-2\sin\text{a}\times1\times\frac12$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=-\sin\text{a}$
View full question & answer→Question 555 Marks
Evaluate the following limit:
Evaluate: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1^4+2^4+3^4+\ \cdots+\text{n}^4}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \cdots+\text{n}^3}{\text{n}^5}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^4+2^4+3^4+\ \cdots+\text{n}^4}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \cdots+\text{n}^3}{\text{n}^5}$
$=\ \lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(1+\text{n})(1+2\text{n})\big(-1+3\text{n}+3\text{n}^2\big)}{30}}{\text{n}^5}-\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)^2}{\text{n}^5}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}^5}\Bigg(\frac{\text{n}^2\big(\text{n}^2+2\text{n}+1\big)}{4}\Bigg)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\frac{1}{\text{n}}+1\big)\big(\frac{1}{\text{n}}+2\big)\big(-\frac{1}{\text{n}^2}+\frac{3}{\text{n}}+3\big)}{30}\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(\frac{1}{\text{n}}+\frac{2}{\text{n}^2}+\frac{1}{\text{n}^3}\Big)}{4}$
$=\frac{1\times2\times3}{30}-0$
$=\frac15$
View full question & answer→Question 565 Marks
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}$
Rationalising the numerator and the denominator:
$=\lim\limits_{\text{x}\rightarrow\infty}\Bigg[\frac{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)}{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}\times\frac{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}\times\frac{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)}{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)}\Bigg]$
$=\lim\limits_{\text{x}\rightarrow\infty}\Bigg[\frac{\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)\big(\sqrt{\text{x}^2+\text{a}^2}-\sqrt{\text{x}^2+\text{b}^2}\big)\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)}{\big(\sqrt{\text{x}^2+\text{c}^2}-\sqrt{\text{x}^2+\text{d}^2}\big)\big(\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}\big)\big(\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}\big)}\Bigg]$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\text{x}^2+\text{a}^2\big)-\big(\text{x}^2+\text{b}^2\big)}{\big(\text{x}^2+\text{c}^2\big)-\big(\text{x}^2+\text{d}^2\big)}\times\bigg(\frac{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}\bigg)$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\bigg(\frac{\sqrt{\text{x}^2+\text{c}^2}+\sqrt{\text{x}^2+\text{d}^2}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2+\text{b}^2}}\bigg)$
Dividing the numerator and the denominator by x:
$=\lim\limits_{\text{x}\rightarrow{\infty}}\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\begin{pmatrix}\frac{\sqrt{1+\frac{\text{a}^2}{\text{x}^2}}+\sqrt{1+\frac{\text{a}^2}{\text{x}^2}}}{\sqrt{1+\frac{1}{\text{a}^2}}+\sqrt{1+\frac{\text{b}^2}{\text{a}^2}}}\end{pmatrix}$
$\text{As x}\rightarrow\infty,\frac{1}{\text{x}},\frac{1}{\text{x}^2}\rightarrow0$
$=\Big(\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}\Big)\Big(\frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}\Big)$
$=\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}$
View full question & answer→Question 575 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2+1-\cos\text{x}}{\text{x}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2+1-\cos\text{x}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^2+2\sin^2\frac{\text{x}}{2}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^2\Bigg[1+2\bigg(\frac{\sin\frac{\text{x}}{2}}{\text{x}}\bigg)^2\Bigg]}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1+2\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2\times\frac{1}{4}}{\frac{\sin\text{x}}{\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1+\lim\limits_{\text{x} \rightarrow0}2\Bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2\times\frac{1}{4}}{\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}}$
$\frac{1+2\times1\times\frac{1}{4}}{1}=\frac{1+\frac{1}{2}}{1}$
$=\frac{3}{2}$
View full question & answer→Question 585 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=9,$ find all possible value of a.
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=9\ \cdots{\text{(i)}}$
$\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}$
$=9(\text{a})^{9-1}$
$=9\text{a}^{8}$
It is given that $9^\text{a}=9$ [From (i)]
$\Rightarrow\text{a}^{8}=\frac99=1$
$\Rightarrow\text{a}^4=1$
$\text{a}^2=1$
$\text{a}=\pm1$
$\Rightarrow \text{a} = 1 \text{ and a} = -1$
View full question & answer→Question 595 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{8}}\frac{\cot4\text{x}-\cos4\text{x}}{(\pi-8\text{x})^3}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{8}}\frac{\cot4\text{x}-\cos4\text{x}}{(\pi-8\text{x})^3}$
When $\text{x}\rightarrow\frac\pi8,\frac\pi8-\text{x}\rightarrow0,$ let $\frac{\pi}{8}-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi8-\text{y}\big)-\cos4\big(\frac\pi8-\text{y}\big)}{(8)^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi2-4\text{y}\big)-\cos4\big(\frac\pi2-4\text{y}\big)}{(8)^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan4\text{y}-\sin4\text{y}}{(8)^2\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{\sin4\text{y}}{\cos4\text{y}}-\sin4\text{y}}{8^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}-\sin4\text{y}\cos4\text{y}}{\cos4\text{y}\times\text{y}^3\times8^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}\big(2\sin^22\text{y}\big)}{\cos4\text{y}\times\text{y}^3\times8^3}$
$=\frac{2}{8^3}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{\text{y}}\times\frac{\sin^22\text{y}}{\text{y}^2}\times\frac{1}{\cos4\text{y}}$
$=\frac{2}{8^3}\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{4\text{y}}\times4\Big)\times\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\cos4\text{y}}$
$=\frac{2}{8^2}(1\times4)\times(1)\times4\times\frac11$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\cos\theta=1\Big]$
$=\frac{2\times4\times4}{8\times8\times8}$
$\frac{1}{16}$
View full question & answer→Question 605 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big\}^{\frac{3\text{x}-2}{3\text{x}+2}}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big\}^{\frac{3\text{x}-2}{3\text{x}+2}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow\infty}\Big\{\Big({\frac{3\text{x}-2}{3\text{x}+2}}\Big)\text{In}\Big(\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big)\Big\}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow\infty}\Bigg\{\Bigg(\frac{3-\frac{2}{\text{x}}}{3+\frac{2}{\text{x}}}\Bigg)\Big(\text{In}\Big(\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big)\Big)\Bigg\}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow\infty}\Bigg\{\Bigg(\frac{3-\frac{2}{\text{x}}}{3+\frac{2}{\text{x}}}\Bigg)\begin{pmatrix}\text{In}\begin{pmatrix}\frac{1+\frac{2}{\text{x}}+\frac{3}{\text{x}^2}}{2+\frac{1}{\text{x}}+\frac{5}{\text{x}^2}}\end{pmatrix}\end{pmatrix}\Bigg\}}$
$=\text{e}^{1.\text{In}\big(\frac{1}{2}\big)}$
$=\frac{1}{2}$
View full question & answer→Question 615 Marks
Evaluate the following limit:
$\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1},\lim\limits_{\text{x}\rightarrow0}\text{ f(x)}=1 $ and $\lim\limits_{\text{x}\rightarrow\infty}\text{f(x)}=1,$ then prove that $\text{f}(-2)=\text{f}(2)=1.$
Answer$\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}$
Also $\lim\limits_{\text{x}\rightarrow0}\text{ f(x)}=1\cdots{(\text{i})}$ [Given]
$\Rightarrow\ \lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1$
$\Rightarrow\ \frac{\lim\limits_{\text{x}\rightarrow0}\text{ax}^2+\text{b}}{\lim\limits_{\text{x}\rightarrow0}\text{x}^2+1}=1$
$\Rightarrow\ \text{b}=1$
Also, it is given that $\lim\limits_{\text{x}\rightarrow\infty}\text{f(x)}=1$
$\therefore\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1\ \cdots(\text{ii})$
$\Rightarrow\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1$
$\Rightarrow\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}+\frac{1}{\text{x}^2}}{1+\frac{1}{\text{x}^2}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$\Rightarrow\text{ a}=1$
Thus, $\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=\frac{\text{x}^2+1}{\text{x}^2+1}=1$ [From (ii)]
$\text{f}(-2)=1$
$\text{f}(2)=1$
$\text{f}(-2)=1=\text{f}(2)$
Hence, proved.
View full question & answer→Question 625 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{5}-\text{a}^5}{\text{x}-\text{a}}=405,$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{5}-\text{a}^5}{\text{x}-\text{a}}=405\ \cdots{\text{(i)}}$
$\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{5}-\text{a}^5}{\text{x}-\text{a}}$
$=5(\text{a})^{5-1}$
$=5\text{a}^{4}$
It is given that $5\text{a}^4=405$
$\Rightarrow5\text{a}^4=405$
$\text{a}^4=\frac{405}{5}=81$
$\text{a}^4=(3)^4,\text{a}^2=9$
$\text{a}=\pm3$
$\Rightarrow\text{a}=3 \text{ and }\text{a}=-3$
View full question & answer→Question 635 Marks
Show that $\lim\limits_{\text{x}\rightarrow0}\ \sin\frac{1}{\text{x}}$ does not exist.
Answer$\lim\limits_{\text{x}\rightarrow0^-}\sin\frac{1}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\sin\frac{1}{0-\text{h}}=-\lim\limits_{\text{h}\rightarrow0}\ \sin\frac{1}{\text{h}}$
= - (Anoscillating number which ascillates between - 1 and 1)
So, $\lim\limits_{\text{x}\rightarrow0^-}\ \sin\frac{1}{\text{x}}$ does not exist.
Similarly, $\lim\limits_{\text{x}\rightarrow0^+}\ \sin\frac{1}{\text{x}}$ does not exist
$\lim\limits_{\text{x}\rightarrow0}\ \sin\frac{1}{\text{x}}$ does not exist.
View full question & answer→Question 645 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2\tan2\text{x}}{\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2\tan2\text{x}}{\tan\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\big(\frac{\text{x}^2}{2\text{x}}-\frac{\tan2\text{x}}{2\text{x}}\big)\times2\text{x}}{\frac{\tan\text{x}}{\text{x}}\times\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\big(\frac{\text{x}}{2}-\frac{\tan2\text{x}}{2\text{x}}\big)}{\frac{\tan\text{x}}{\text{x}}}$
$=2\Big(\frac{0-1}{1}\Big)$
$=-2$
View full question & answer→Question 655 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\big(\frac\pi2-\text{x}\big)\sin\text{x}-2\cos\text{x}}{\big(\frac\pi2-\text{x}\big)+\cot\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\big(\frac\pi2-\text{x}\big)\sin\text{x}-2\cos\text{x}}{\big(\frac\pi2-\text{x}\big)+\cot\text{x}}$
If $\text{x}\rightarrow\frac{\pi}{2},$ then $\frac\pi2-\text{x}\rightarrow0$ let $\frac\pi2-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big(\text{y}\sin\big(\frac{\pi}{2}-\text{y}\big)-2\cos\big(\frac{\pi}{2}-\text{y}\big)\Big)}{\text{y}+\cot\big(\frac\pi2-\text{y}\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\Big(\frac{\text{y}\cos\text{y}-2\sin\text{y}}{1+\tan\text{y}}\Big)$
$=\lim\limits_{\text{y}\rightarrow{0}}\Bigg(\frac{\cos\text{y}-2\frac{\sin\text{y}}{\text{y}}}{1+\frac{\tan\text{y}}{\text{y}}}\Bigg)$
$=\frac{\lim\limits_{\text{y}\rightarrow{0}}\cos\text{y}-2\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}}{1+\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan\text{y}}{\text{y}}}$ $\Big[\because\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}=1,\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}=1\Big]$
$=\frac{1-2}{1+1}=\frac{-1}{2}$
$=-\frac12$
View full question & answer→Question 665 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}}$
Answer$\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}-5}{\big(\sqrt{6\text{x}-5}-\sqrt{4\text{x}+5}\big)}\times\frac{\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{(6\text{x}-5)-(4\text{x}+5)}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{2\text{x}-10}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{(\text{x}-5)\big(\sqrt{6\text{x}-5}+\sqrt{4\text{x}+5}\big)}{2(\text{x}-5)}$
$=\frac{\sqrt{6(5)-5}+\sqrt{4(5)+5}}{2}$
$=\frac{\sqrt{25}+\sqrt{25}}{2}$
$=\frac{5+5}{2}=5$
View full question & answer→Question 675 Marks
Evaluate the following limit:
$\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$
Answer$\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin(\text{a}+\text{h})-\text{a}^2\sin\text{a}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2(\sin\text{a}\cos\text{h})-\text{a}^2\sin\text{a}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{a}^2+2\text{ah}+\text{h}\big)(\sin\text{a}\cos\text{h})-\text{a}^2\sin\text{a}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin\text{a}(\cos-1)+2\text{ah}\sin\text{a}\cos\text{h}+\text{h}^2\sin\text{a}\cos\text{h}+(\text{a}+\text{h})^2\cos\text{a}\sin\text{h}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}^2\sin\text{a}(\cos-1)}{\text{h}}+\lim\limits_{\text{h}\rightarrow0}\frac{2\text{ah}\sin\text{a}\cos\text{h}}{\text{h}}\\\ +\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2\sin\text{a}\cos\text{h}}{\text{h}}+\lim\limits_{\text{h}\rightarrow0}\frac{(\text{a}+\text{h})^2\sin\text{a}\cos\text{h}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{a}^2\sin\text{a}\sin^2\big(\frac{\text{h}}{2}\big)}{\frac{\text{h}}{2}}+2\text{a}\sin\text{a}+0+\text{a}^2\cos\text{a}$
$=0+2\text{a}\sin\text{a}+\text{a}^2\cos\text{a}$
$=2\text{a}\sin\text{a}+\text{a}^2\cos\text{a}$
View full question & answer→Question 685 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\{\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\text{x}\}}{\cos^2\beta\text{x}-\cos^2\alpha\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\{\sin(\alpha+\beta)\text{x}+\sin(\alpha-\beta)\text{x}+\sin2\alpha\text{x}\}}{\cos^2\beta\text{x}-\cos^2\alpha\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\Big\{2\sin\frac{(\alpha+\beta+\alpha-\beta)}{2}\times\cos\frac{(\alpha+\beta-\alpha+\beta)}{2}\times+2\sin\alpha\cos\alpha\text{x}\Big\}}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big\{2\sin\alpha\text{x}\cos\beta\text{x}+2\sin\alpha\text{x}\cos\alpha\text{x}\big\}}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}(\cos\beta\text{x}+\cos\alpha\text{x})}{(\cos\beta\text{x}-\cos\alpha\text{x})(\cos\beta\text{x}+\cos\alpha\text{x})}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{(\cos\beta\text{x}-\cos\alpha\text{x})}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{\Big(1-2\sin^2\big(\frac{\beta\text{x}}{2}\big)-1+2\sin^2\big(\frac{\alpha\text{x}}{2}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\alpha\text{x}}{2\sin^2\big(\frac{\alpha\text{x}}{2}\big)-2\sin^2\big(\frac{\beta\text{x}}{2}\big)}$
$=\frac{2\alpha}{\alpha^2-\beta^2}$
View full question & answer→Question 695 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\text{cosec}^2\text{x}}{1-\cot\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\text{cosec}^2\text{x}}{1-\cot\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-\big(1+\cot^2\text{x}\big)}{1-\cot\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{2-1-\cot^2\text{x}}{1-\cot\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{1-\cot^2\text{x}}{1-\cot\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{(1-\cot\text{x})(1+\cot\text{x})}{(1-\cot\text{x})}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}(1+\cot\text{x})$
$=1+\cot\frac{\pi}{4}$
$=1+1$
$=2$
View full question & answer→Question 705 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+3\text{x}}-\sqrt{1-3\text{x}}\big)}{\text{x}}\times\frac{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(1+3\text{x})-(1-3\text{x})}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{6\text{x}}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{6}{\big(\sqrt{1+3\text{x}}+\sqrt{1-3\text{x}}\big)}$
$=\frac{6}{\sqrt{1}+\sqrt{1}}$
$=\frac62$
$=3$
View full question & answer→Question 715 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^7-2\text{x}^5+1}{\text{x}^3-3\text{x}^2+2}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^7-2\text{x}^5+1}{\text{x}^3-3\text{x}^2+2}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\big(\text{x}^6+\text{x}^5-\text{x}^4-\text{x}^3-\text{x}^2-\text{x}-1\big)}{(\text{x}-1)\big(\text{x}^2-2\text{x}-2\big)}$
$= \lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^6+\text{x}^5-\text{x}^4-\text{x}^3-\text{x}^2-\text{x}-1\big)}{\big(\text{x}^2-2\text{x}-2\big)}$
$=\frac{(1+1-1-1-1-1-1)}{(1-2-2)}$
$=\frac{-3}{-3}$
$=1$
View full question & answer→Question 725 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}\big(3\sin\text{x}-4\sin^3\text{x}\big)}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{4\sin^3\text{x}}{\text{x}^3}$
$=4\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)^3$
$=4\times1$
$=4$
View full question & answer→Question 735 Marks
Evaluate the following limit:
If $\lim\limits_{\text{x}\rightarrow0}{\text{ kx cosec x}}=\lim\limits_{\text{x}\rightarrow0}\text{ x cosec kx,}{}$ find k.
Answer$\lim\limits_{\text{x}\rightarrow0}{\text{ kx cosec x}}=\lim\limits_{\text{x}\rightarrow0}\text{ x cosec kx,}{}$
$\lim\limits_{\text{x}\rightarrow0}\text{ kx}\frac{1}{\sin\text{x}}=\lim\limits_{\text{x}\rightarrow0}\text{x}\frac{1}{\sin\text{kx}}$
$\text{k}\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{x}}{\sin\text{x}}\Big)=\frac{1}{\text{k}}\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{kx}}{\sin\text{kx}}\Big)$
$\text{k}=\frac{1}{\text{k}}$
$\text{k}^2=1$
$\text{k}=\pm1$
View full question & answer→Question 745 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}},\text{x}>1$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}\times\frac{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}\times\frac{\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^-1}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\big[\big(\text{x}^2-1\big)-(\text{x}-1)\big]\times\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\text{x}\big)\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\sqrt{\text{x}^2-1}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}{(\text{x}+1)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}-1\big)}$
$=\frac{\sqrt{2}}{2\big(\sqrt{2}-1\big)}$
$=\frac{\sqrt{2}}{2\times\big(\sqrt{2}-1\big)}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}$
$=\frac{\sqrt{2}+1}{\sqrt{2}}$
View full question & answer→Question 755 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}2\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{4}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}2\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{4}\big)}=\lim\limits_{\text{h}\rightarrow{0}}\frac{1\sin\big(\frac{\pi+\text{h}}{2}\big)}{\cos\big(\frac{\pi+\text{h}}{2}\big)\big(\cos\big(\frac{\pi+\text{h}}{2}\big)-\sin\big(\frac{\pi+\text{h}}{2}\big)\big)}$
$=\lim\limits_{\text{h}\rightarrow{0}}\frac{1-\cos\big(\frac{\text{h}}{2}\big)}{-\sin\big(\frac{\text{h}}{2}\big)\big(\frac{1}{\sqrt{2}}\cos\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\sin\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\sin\big(\frac{\text{h}}{4}\big)-\frac{1}{\sqrt{2}}\cos\big(\frac{\text{h}}{4}\big)\big)}$
$=\lim\limits_{\text{h}\rightarrow{0}}\frac{1-\cos\big(\frac{\text{h}}{2}\big)}{\sqrt{2}\sin\big(\frac{\text{h}}{2}\big)\sin\big(\frac{\text{h}}{4}\big)}$
$=\lim\limits_{\text{h}\rightarrow{0}}\frac{2-\sin^2\big(\frac{\text{h}}{4}\big)}{\sqrt{2}\sin\big(\frac{\text{h}}{2}\big)\sin\big(\frac{\text{h}}{4}\big)}$
$=\sqrt{2}\lim\limits_{\text{h}\rightarrow{0}}\frac{\sin\big(\frac{\text{h}}{4}\big)}{\sin\big(\frac{\text{h}}{2}\big)}$
$=\sqrt{2}\lim\limits_{\text{h}\rightarrow{0}}\frac{\frac{\sin\big(\frac{\text{h}}{4}\big)}{\big(\frac{\text{h}}{4}\big)}\times\big(\frac{\text{h}}{4}\big)}{\frac{\sin\big(\frac{\text{h}}{2}\big)}{\big(\frac{\text{h}}{2}\big)}\times\big(\frac{\text{h}}{2}\big)}$
$=\sqrt{2}\times\frac{\frac14}{\frac12}$
$=\frac{1}{\sqrt{2}}$
View full question & answer→Question 765 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-3\text{x}^2+9\text{x}-2\text{ by }\text{x}^3-\text{x}-6$ 
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9}{\text{x}^3-\text{x}-6}=\lim\limits_{\text{x}\rightarrow2}1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-2\text{x}-6\text{x+4}}{\text{x}^3-\text{x}-6}$ $\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}-1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-\text{x}-6\text{ by}\text{ x}-2$ 
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{(\text{x}-2)(\text{x}^2+2\text{x}+3)}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)}{(\text{x}^2-2\text{x}+3)}$ $=1+\frac{3\times2-2}{2^2+2\times2+3}$ $=1+\frac{4}{11}$ $=\frac{15}{11}$ View full question & answer→Question 775 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{5+\frac{4\sin3\text{x}}{\text{x}}}{\frac{4\sin2\text{x}}{\text{x}}+7}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}5+4\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3}{4\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2+7}$
$=\frac{5+4\times1\times3}{4\times2+7}$
$=\frac{5+12}{8+7}$
$=\frac{17}{15}$
View full question & answer→Question 785 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\text{cosec}^2\text{x}-2}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}+1-2}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{\cot^2\text{x}-1}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}\frac{(\cot\text{x}-1)(\cot\text{x}+1)}{\cot\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}}(\cot\text{x}+1)$
$=\cot\frac{\pi}{4}+1$
$=1+1$
$=2$
View full question & answer→Question 795 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{2}+1\big)}{\text{x}^2-2}$
Answer$\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{2}+1\big)}{\text{x}^2-2}$ $=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\Big(\sqrt{\big(\sqrt{2}+1\big)^2}\Big)}{\text{x}^2-2}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\big(\sqrt{3+2\sqrt{2}}\big)}{\text{x}^2-2}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\big(\sqrt{3+2\sqrt{2}}\big)}{\text{x}^2-2}\times\frac{\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)}{\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{{3+2\text{x}}-{3-2\sqrt{2}}}{\big(\text{x}^2-2\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{2\big(\text{x}-\sqrt{2}\big)}{\big(\text{x}^2-2\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{2}{\big(\text{x}^2-\sqrt{2}\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\frac{2}{\big(\sqrt{2}+\sqrt{2}\big)\Big(\sqrt{3+2\sqrt{2}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\frac{2}{\big(2\sqrt{2}\big)\Big(2\sqrt{3+2\sqrt{2}}\Big)}$
$=\frac{1}{\big(2\sqrt{2}\big)\Big(\sqrt{3+2\sqrt{2}}\Big)}$
$=\frac{1}{\big(2\sqrt{2}\big)\Big(\sqrt{2}+1\Big)}$
$=\frac{\big(\sqrt{2}-1\big)}{\big(2\sqrt{2}\big)}$
View full question & answer→Question 805 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}$ Dividing $\text{x}^4-3\text{x}^3+2\text{ by }\text{x}^3-5\text{x}^2+3\text{x}+1$ 
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-3\text{x}^3+2}{\text{x}^3-5\text{x}^2+3\text{x}+1}=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ Dividing $\text{x}^3-5\text{x}^2+3\text{x}+1\text{ by }\text{x}-1$ 
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}^2-7\text{x}}{\text{x}^3-5\text{x}^3+3\text{x}+1}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}(\text{x}-1)}{(\text{x}-1)(\text{x}^2-4\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\text{ x}+2+\lim\limits_{\text{x}\rightarrow1}\frac{7\text{x}}{\big(\text{x}^2-4\text{x}-1\big)}$ $=1+2+\frac{7}{(1-4-1)}$ $=3-\frac{7}{4}$ $=\frac{12-7}{4}$ $=\frac{5}{4}$ View full question & answer→Question 815 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}}-1\big)}{\text{x}}\times\frac{\big(\sqrt{1+\text{x}}+1\big)}{\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(1+\text{x}-1)}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\big(\sqrt{1+\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{1+\text{x}}+1\big)}$
$=\frac{1}{\sqrt{1}+1}=\frac12$
View full question & answer→Question 825 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \cdots+\text{n}^2}{\text{n}^3}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \cdots+\text{n}^2}{\text{n}^3}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac16\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{\text{n}^3}$ $\Big[1^2+2^2+3^2+\ \cdots+\text{n}^2=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+1)\Big]$
$=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\text{n}^2+\text{n}\big)(2\text{n}+1)}{\text{n}^3}$
$=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(2\text{n}^3+\text{n}^2+2\text{n}^2+\text{n}\big)}{\text{n}^3}$
$=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\big(2\text{n}^2+3\text{n}^2+\text{n}\big)}{\text{n}^3}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(2+\frac{3}{\text{n}}+\frac{1}{\text{n}^2}\Big)}{1}$
$=\frac16\frac{(2+0+0)}{1}=\frac{1}{3}$
View full question & answer→Question 835 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{5}+\sqrt{2}\big)}{\text{x}^2-10}$
Answer$\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{5}+\sqrt{2}\big)}{\text{x}^2-10}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\Big(\sqrt{\big(\sqrt{5}+\sqrt{2}\big)^2}\Big)}{\text{x}^2-10}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{7+2\sqrt{10}}\big)}{\text{x}^2-10}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{\sqrt{7+2\text{x}}-\big(\sqrt{7+2\sqrt{10}}\big)}{\text{x}^2-10}\times\frac{\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)}{\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{{7+2\text{x}}-{7-2\sqrt{10}}}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{2\big(\text{x}-\sqrt{10}\big)}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{2}{\big(\text{x}-\sqrt{10}\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$
$=\frac{2}{\big(\sqrt{10}+\sqrt{10}\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$
$=\frac{2}{\big(2\sqrt{10}\big)\Big(2\sqrt{7+2\sqrt{10}}\Big)}$
$=\frac{1}{\big(2\sqrt{10}\big)\Big(\sqrt{7+2\sqrt{10}}\Big)}$
$=\frac{1}{\big(2\sqrt{10}\big)\Big(\sqrt{5}+\sqrt{2}\Big)}$
$=\frac{\big(\sqrt{5}-\sqrt{2}\big)}{\big(6\sqrt{10}\big)}$
View full question & answer→Question 845 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-\cos\text{dx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-\cos\text{dx}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{{\Big(-2\sin\big(\frac{\text{a}+\text{b}}{2}\big)\times\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\Big)}}{-2\sin\big(\frac{\text{c}+\text{d}}{2}\big)\times\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}{\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}}{\lim\limits_{\text{x}\rightarrow0}{\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}}$
$=\frac{\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}{\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}}\times\big(\frac{\text{a}+\text{b}}{2}\big)\text{x}\Bigg)\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}}{\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}}\times\big(\frac{\text{a}-\text{b}}{2}\big)\text{x}\Bigg)}{\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}{\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}}\times\big(\frac{\text{c}+\text{d}}{2}\big)\text{x}\Bigg)\Bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}{\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}}\times\big(\frac{\text{c}-\text{d}}{2}\big)\text{x}\Bigg)}$
$=\frac{(\text{a}+\text{b})(\text{a}-\text{b})}{(\text{c}+\text{d})(\text{c}-\text{d})}$ $\Big[\because\ \lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{\text{a}^2-\text{b}^2}{\text{c}^2-\text{d}^2}$
View full question & answer→Question 855 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{4\text{x}^2-7\text{x}}+2\text{x}\big)$
Answer$\lim\limits_{\text{x}\rightarrow-\infty}\big(\sqrt{4\text{x}^2-7\text{x}}+2\text{x}\big)$
Substitute y = -x
$=\lim\limits_{\text{y}\rightarrow\infty}\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(\sqrt{4\text{y}^2+7\text{y}}-2\text{y}\Big)\Big(\sqrt{4\text{y}^2+7\text{y}}+2\text{y}\Big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{\Big(4\text{y}^2+7\text{y}+4\text{y}^2\big)}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{(7\text{y})}{\sqrt{4\text{y}^2+7\text{y}}+2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow\infty}\frac{7}{\sqrt{4+\frac{7}{\text{y}}}+2}$
$=\frac{7}{2+2}=\frac{7}{4}$
View full question & answer→Question 865 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}-\sin2\text{x}}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin2\text{x}}{\cos2\text{x}}-\sin2\text{x}}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(\frac{1}{\cos2\text{x}}-1\big)}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(1-\cos2\text{x})}{\text{x}^2\cos2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}\big(2\sin^2\text{x}\big)}{\text{x}^3\cos2\text{x}}$
$=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\big)\Big(\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{\text{x}^2}\Big)}{\big(\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}\big)}$
$=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{x}}\times2\big)\Big(\lim\limits_{\text{x}\rightarrow0}2\big(\frac{\sin\text{x}}{\text{x}}\big)^2\Big)}{\lim\limits_{\text{x}\rightarrow0}\cos2\text{x}}$
$=\frac{(2\times1)(2\times1)}{1}$
$=4$
View full question & answer→Question 875 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{1+\text{cosec}^3\text{x}}{\cot^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{1+\text{cosec}^3\text{x}}{\cot^2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{(1+\text{cosec}\text{x})\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{ x}\big)}{\big(\text{cosec}^2\text{x}-1\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{(\text{cosec}\text{x}+1)\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{x}\big)}{(\text{cosecx}-1)\big(\text{cosecx+1}\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{3\pi}{2}}\frac{\big(1+\text{cosec}^2\text{x}-\text{cosec}\text{ x}\big)}{(\text{cosec }\text{x}-1)}$
$=\frac{1+\text{cosec}^2\frac{3\pi}{2}-\text{cosec}\frac{3\pi}{2}}{\text{cosec}\frac{3\pi}{2}-1}$
$=\frac{1+(-1)^2-(-1)}{(-1)-1}$ $\Big[\therefore\text{cosec}\frac{3\pi}{2}=-1\Big]$
$=\frac{1+1+1}{-2}$
$=\frac{-3}{2}$
View full question & answer→Question 885 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos5\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos5\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(-2\sin\big(\frac{3\text{x}+5\text{x}}{2}\big)\sin\big(\frac{3\text{x}-5\text{x}}{2}\big)\Big)}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{-2\sin4\text{x}\sin(-\text{x})}{\text{x}^2}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin4\text{x}\sin\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin4\text{x}}{\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$
$=2\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin4\text{x}}{4\text{x}}\times4\Big)\times\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=8$
View full question & answer→Question 895 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}-2\text{x}}{3\text{x}-\sin^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}-2\text{x}}{3\text{x}-\sin^2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\tan3\text{x}}{\text{x}}-\frac{\tan2\text{x}}{\text{x}}}{\frac{3\text{x}}{\text{x}}-\frac{\sin^2\text{x}}{\text{x}}}$
$=\frac{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\times3\Big)-\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}}{2\text{x}}\times2\Big)}{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{3\text{x}}{2}\Big)-\lim\limits_{\text{x}\rightarrow0}\frac{(\sin\text{x})^2}{\text{x}}}$
$=\frac{3-2}{3-\big(\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{3-2}{3-0}=\frac13$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac13$
View full question & answer→Question 905 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{{(\text{n}+2)!}+{(\text{n}+1)!}}{{(\text{n}+2)!}+{(\text{n}+1)!}}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{{(\text{n}+2)!}+{(\text{n}+1)!}}{{(\text{n}+2)!}+{(\text{n}+1)!}}$
We know that (n + 2) = (n + 2)(n + 1)!
$\Rightarrow\lim\limits_{\text{n}\rightarrow\infty}\frac{(\text{n}+2)(\text{n}+1)!+(\text{n}+1)!}{(\text{n}+2)(\text{n}+1)!-(\text{n}+1)!}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{(\text{n}+1)!\big[(\text{n}+2)+1\big]}{(\text{n}+1)\big[(\text{n}+2)-1\big]}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}+3}{\text{n}+1}$ $\Big[\frac\infty\infty\text{ from}\Big]$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\frac{1+\frac{3}{\text{n}}}{1+\frac{1}{\text{n}}}$
$=\frac{1+0}{1+0}$
$=1$
View full question & answer→Question 915 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$
As$\text{ x}\rightarrow\pi,\text{x}-\pi\rightarrow0,$let $\text{ x }-\pi=\text{y}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\tan^2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$
$=\frac{\lim\limits_{\text{y}\rightarrow0}2\sin^2\frac{\text{y}}{2}}{{\lim\limits_{\text{y}\rightarrow0}\tan^2\text{y}}}$
$=\frac{2\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}}{\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\bigg)\times\text{y}^2}$
$=\frac{2\times1\times\frac{\text{y}^2}{4}}{1\times\text{y}^2}$ $\begin{bmatrix}\therefore\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\\\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1 \end{bmatrix}$
$=2\times1\times\frac{1}{4}$
$=\frac{1}{2}$
View full question & answer→Question 925 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$
As x → 1, then x - 1 → 0 let x - 1 = y
$=\lim\limits_{{\text{x}-{1\rightarrow0}}}\frac{(\text{x}-1)}{\text{x}\times\sin\pi(\text{x}-1)}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{1}{\frac{(\text{y}+1)\sin(\pi\text{y})}{\text{y}}}$
$=\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow0}(\text{y}+1)\Big)\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\pi\text{y}}{\text{y}\times\pi}\times\pi\Big)}$
$=\frac{1}{(1)(1\times\pi)}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{1}{\pi}$
View full question & answer→Question 935 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)}{\text{x}-1}\times\frac{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{((5\text{x}-4)-\text{x)}}{(\text{x}-1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=4\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)}{(\text{x}-1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=4\lim\limits_{\text{x}\rightarrow1}\frac{1}{\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=4\times\frac{1}{\sqrt{5-4}+\sqrt{1}}$
$=4\times\frac{1}{\sqrt{5-4}+\sqrt{1}}$
$=4\times\frac{1}{\sqrt{1}+\sqrt{1}}$
$=\frac{4}{2}=2$
View full question & answer→Question 945 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}$
$=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{\sqrt{2+\cos\text{x}}-1}{({\pi-\text{x}})^2}\times\frac{\sqrt{2+\cos\text{x}}+1}{{\sqrt{2+\cos\text{x}}+1}}$
$=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{(2+\cos\text{x})-1}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}$
Let $\pi-\text{x}=\text{y},\text{x}\rightarrow\pi,\text{y}\rightarrow0$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\pi}}\frac{1+\cos\text{x}}{(\pi-\text{x})^2\big(\sqrt{2+\cos\text{x}}+1\big)}=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos\text{x}(\pi-\text{y})}{\text{y}^2\big(\sqrt{2+\cos(\pi-\text{y})+1}\big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\text{y}^2\sqrt{2-\cos\text{y}+1}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\frac{2\sin^2\text{y}}{2}}{\text{y}^2\sqrt{2-\cos\text{y}}+1}$
$=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{1}{4}\frac{1}{\sqrt{2-\cos\text{y}}+1}$
$=2\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{2}\Big)^2\times\frac{1}{4}\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos\text{y}}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-\cos\text{0}}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{\sqrt{2-1}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{{1}+1}$
$=2\times1\times\frac{1}{4}\times\frac{1}{{2}}$
$=\frac{1}{4}$
View full question & answer→Question 955 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(\cos3\text{x}-\cos\text{x})}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}(\cos3\text{x}-\cos\text{x})}{\text{x}^3}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}\Big(-2\sin\big(\frac{3\text{x}+\text{x}}{2}\big)\sin\big(\frac{3\text{x}-\text{x}}{2}\big)\Big)}{\text{x}^3}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}(-2\sin2\text{x}\sin\text{x})}{\text{x}^3}$
$=\frac{-2\lim\limits_{\text{x} \rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x} \rightarrow0}\sin2\text{x}\times\lim\limits_{\text{x} \rightarrow0}\sin\text{x}}{\text{x}^3}$
$=-2\Big(\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)\times\Big(2\lim\limits_{\text{x} \rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\Big)\times\Big(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)$
$=-2(1\times2)\times(2)\times(1)$
$=-8$
View full question & answer→Question 965 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$
Answer$\lim\limits_{\text{x}\rightarrow{\infty}}{\sqrt{\text{x}^2+\text{cx}}-\text{x}}{}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\Bigg(\big(\sqrt{\text{x}^2+\text{cx}}-\text{x}\big)\frac{\big(\sqrt{\text{x}^2+\text{cx}}+\text{x}\big)}{\sqrt{\text{x}^2+\text{cx}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\big(\text{x}^2+\text{cx}-\text{x}^2\big)}{\sqrt{\text{x}^2+\text{cx}+\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{cx}}{\sqrt{\text{x}^2+\text{cx}+\text{x}}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{c}}{\sqrt{1+\frac{\text{c}}{\text{x}}+1}}$
$=\frac{\text{c}}{1+1}=\frac{\text{c}}{2}$
View full question & answer→Question 975 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^2-1}$
Answer $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{5\text{x}-4}-\sqrt{\text{x}}}{\text{x}^2-1}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{5\text{x}-4}-\sqrt{\text{x}}\big)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{((5\text{x}-4)-\text{x})}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4(\text{x}-1)}{(\text{x}-1)(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{4}{(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{\text{x}}\big)}$
$=\frac{4}{(\text{x}+1)\big(\sqrt{5\text{x}-4}+\sqrt{1}\big)}$
$=\frac{4}{2(1+1)}$
$=\frac{4}{4}=1$
View full question & answer→Question 985 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}{\Big(\frac{1}{\text{n}^2}+\frac{2}{\text{n}^2}+\frac{3}{\text{n}^2}+\ \cdots+\frac{\text{n}-1}{\text{n}^2}}{}\Big)$
Answer$\lim\limits_{\text{n}\rightarrow\infty}{\Big(\frac{1}{\text{n}^2}+\frac{2}{\text{n}^2}+\frac{3}{\text{n}^2}+\ \cdots+\frac{\text{n}-1}{\text{n}^2}}{}\Big)$ $=\lim\limits_{\text{n}\rightarrow\infty}\Big(\frac{1+2+3+\ \cdots+(\text{n}-1)}{{n}^2}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\text{n}+1\big)(\text{n})}{2\times\text{n}^2}$ $\Big[1+2+3+\ \cdots+({\text{n}}-1)=\frac{({\text{n}}-1)({\text{n}})}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}^2-\text{n}}{2\text{n}^2}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\frac{1-\frac{1}{\text{n}}}{2}$
$=\frac{1-0}{2}=\frac12$
$=\frac{1}{2}$
View full question & answer→Question 995 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$
Answer $\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-\sqrt{\text{x}}}{\sqrt{\text{x}}-1}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\sqrt{\text{x}}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}{\big(\sqrt{\text{x}}-1\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-\text{x}}{\big(\sqrt{\text{x}-1}\big)\big({\text{x}^2}+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\text{x}^3-1\big)}{\big(\sqrt{\text{x}-1}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\big(\text{x}^2+1+\text{x}\big)}{\big(\sqrt{\text{x}-1}\big)\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}+1}\big)\big({\text{x}^2+1+\text{x}}\big)}{\big(\text{x}^2+\sqrt{\text{x}}\big)}$
$=\frac{1(1+1)(1+1+1)}{1+1}$
$=\frac62$
$=3$
View full question & answer→Question 1005 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$
Answer $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$ [Dividing numerator and denominator by x - a]
$=\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^\frac{3}{4}-\text{a}^\frac{3}{4}}{\text{x}-\text{a}}}$
Applying the formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ in numerator and $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{m}}-\text{a}^{\text{m}}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator respectively
Here, $\text{n}=\frac{2}{3},\text{m}=\frac{3}{4}$
$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\frac{2}{3}}-\text{a}^{\frac{2}{3}}}{\text{x}-\text{a}}}{\lim\limits_{\text{x}\rightarrow{\text{x}}}\frac{\text{x}^{\frac{3}{4}}-\text{a}^{\frac{3}{4}}}{\text{x}-\text{a}}}=\frac{\frac{2}{3}(\text{a})^{\frac{2}{3}-1}}{\frac{3}{4}(\text{a})^{\frac{3}{4}-1}}$
$=\frac{8}{9}\text{a}^{\frac{-1}{3}+\frac{1}{4}}$
$=\frac89\text{a}^{\frac{-1}{12}}$
View full question & answer→