Question 513 Marks
Evaluate the following limit:
$\lim\limits_{\theta\rightarrow0}\frac{\sin4\theta}{\tan3\theta}$
Answer$\lim\limits_{\theta\rightarrow0}\frac{\sin4\theta}{\tan3\theta}$
$=\frac{\lim\limits_{\theta\rightarrow0}\sin4\theta}{\lim\limits_{\theta\rightarrow0}\tan3\theta}$
$=\frac{\Big(\lim\limits_{\theta\rightarrow0}\frac{\sin4\theta}{4\theta}\Big)\times4\theta}{\Big(\lim\limits_{\theta\rightarrow0}\frac{\tan3\theta}{3\theta}\Big)\times3\theta}$
$=\frac{1\times4\theta}{1\times3\theta}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac43$
View full question & answer→Question 523 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{m}\text{x}}{\tan\text{n}\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{m}\text{x}}{\tan\text{n}\text{x}}$$=\frac{\lim\limits_{\text{x}\rightarrow0}{\tan\text{m}\text{x}}{}}{\lim\limits_{\text{x}\rightarrow0}{\sin\text{n}\text{x}}}$
$=\frac{\lim\limits_{\text{m}\text{x}\rightarrow0}\frac{\tan\text{m}\text{x}}{\text{m}\text{x}}\times\text{m}\text{x}}{\lim\limits_{\text{n}\text{x}\rightarrow0}\frac{\tan\text{n}\text{x}}{\text{n}\text{x}}\times\text{n}\text{x}}$ $[\because$ if x → 0 then mx → 0 also nx → 0$]$
$=\frac{1\times\text{m}}{1\times\text{n}}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{\text{m}}{\text{n}}$
View full question & answer→Question 533 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{-\frac{1}{2}}}\frac{8\text{x}^{3}+1}{2\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow{-\frac{1}{2}}}\frac{8\text{x}^{3}+1}{2\text{x}+1}$$=\frac82\lim\limits_{\text{x}\rightarrow{-\frac{1}{2}}}\frac{\text{x}^{3}+\big(\frac12\big)^3}{\text{x}+\frac12}$
$=4\lim\limits_{\text{x}\rightarrow{-\frac{1}{2}}}\frac{\text{x}^{3}+\big(\frac12\big)^3}{\text{x}-\big(-\frac12\big)}$
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, n = 3, $\text{a}=\frac{-1}{2}$
$=4\lim\limits_{\text{x}\rightarrow{-\frac{1}{2}}}\frac{\text{x}^{3}+\big(\frac12\big)^3}{\text{x}-\big(-\frac12\big)}=4\times3\Big(-\frac{1}{2}\Big)^{3-1}$
$=4\times3\times\frac14$
$=3$
View full question & answer→Question 543 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4\text{x}-\pi)^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4\text{x}-\pi)^2}$
$\text{x}\rightarrow\frac{\pi}{4}$ then $\text{x}-\frac\pi4\rightarrow0,$ also $4\text{x}-\pi\rightarrow0$ let $\text{x}-\frac{\pi}{4}\rightarrow\text{y}$
$=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{(4)^2\big(\text{x}-\frac{\pi}{4}\big)^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\cos\big({\text{y}+\frac\pi4\big)-\sin\big({\text{y}+\frac\pi4\big)}}}{16\times\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\big(\cos\text{y}\times\frac{1}{\sqrt{2}}-\sin\text{y}\times\frac{1}{\sqrt{2}}\big)-\big(\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}\big)}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}(\cos\text{y}-\sin\text{y})-\frac{1}{\sqrt{2}}(\sin\text{y}+\cos\text{y})}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}\big[(\cos\text{y}-\sin\text{y})-(\sin\text{y}+\cos\text{y})\big]}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\Big(\sqrt{2}-\frac{1}{\sqrt{2}}\cos\text{y}+\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)}{\text{y}^2}$
$=\frac{1}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\frac{1}{\sqrt{2}}\cos\text{y}}{\text{y}^2}$
$=\frac{\sqrt{2}}{16}\lim\limits_{\text{y}\rightarrow{0}}\frac{{2}\sin^2\frac{\text{y}}{{2}}}{\text{y}^2}$
$=\frac{\sqrt{2}}{8}\lim\limits_{\text{y}\rightarrow{0}}\Bigg(\frac{\sin\frac{\text{y}}{{2}}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14$
$=\frac{1}{16\sqrt{2}}$
View full question & answer→Question 553 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}\cos\text{x}+\sin\text{x}}{3\text{x}^2+\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}\cos\text{x}+\sin\text{x}}{3\text{x}^2+\tan\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{5\cos\text{x}+3\frac{\sin\text{x}}{\text{x}}}{3\text{x}+\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}5\cos\text{x}\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x}\rightarrow0}3\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}5\cos\text{x}+3\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}{3\lim\limits_{\text{x}\rightarrow0}\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{5\times\cos0+3\times1}{3\times0+1}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{5+3}{1}$
$=8$
View full question & answer→Question 563 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}}{\text{x}-3}-\frac{4}{\text{x}^2-2\text{x}}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}}{\text{x}-2}-\frac{4}{\text{x}^2-2\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}}{\text{x}-2}-\frac{4}{\text{x}(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}(\text{x}-4)}{\text{x}(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}^2-4}{\text{x}(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)}{\text{x}}$
$=\frac{2+2}{2}$
$=\frac42$
$=2$
View full question & answer→Question 573 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^3\cot\text{x}}{1-\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^3\cot\text{x}}{1-\cos\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^3}{\tan\text{x}(1-\cos\text{x})}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^3}{\tan\text{x}.2\sin^2\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1}{\frac{\tan\text{x}}{\text{x}}\times\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}}$
$=\frac{1}{\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}\bigg)\times2\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)^2\times\frac14}$
$=\frac{1}{2\times2\times1\times\frac14}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$
$=2$
View full question & answer→Question 583 Marks
Show that $\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}\ne\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{a}]}.$
Answer$\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}$
$=\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{1}=\frac{2}{1}=2$ $\bigg[\because\lim\limits_{\text{x}\rightarrow\text{k}^-}\ [\text{x}]=\text{k}-1\bigg]$
Also,
$\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{x}]}=\lim\limits_{\text{x}\rightarrow2^-}\frac{\text{x}}{3}=\frac23$ $\bigg[\because\lim\limits_{\text{x}\rightarrow\text{k}^+}\ [\text{x}]=\text{k}+1\bigg]$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2^-}\ \frac{\text{x}}{[\text{x}]}\ne\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}}{[\text{x}]}$
View full question & answer→Question 593 Marks
Evaluate $\lim\limits_{\text{x}\rightarrow0}\text{f(x)},$ where $\text{f(x)}=\begin{cases}\frac{|\text{x}|}{\text{x}}, & \text{x} \ne0\\0, &\text{x} = 0\end{cases}.$
Answer$\text{L.H.L}=\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^-}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{|0-\text{h}|}{0-\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{+\text{h}}{-\text{h}}=-1\ \cdots(\text{i})$
And,
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^+}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^+}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{|0+\text{h}|}{0+\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}=1\ \cdots(\text{ii})$
So, $\text{L.H.L}\ne\text{R.H.L}$
$\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.
View full question & answer→Question 603 Marks
Let $\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, & \text{where x} \ne\frac\pi2\\3, & \text{where x} \ne\frac\pi2\end{cases}$ and if $\lim\limits_{\text{x}\rightarrow\frac\pi2}{\text{f(x)}}=\text{f}\Big(\frac\pi2\Big),$ find the value of k.
AnswerWe have,
$\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, & \text{where x} \ne\frac\pi2\\3, & \text{where x} \ne\frac\pi2\end{cases}$
It is given that,
$\lim\limits_{\text{x}\rightarrow\frac\pi2}{\text{f(x)}}={{\text{f}\big(\frac{\pi}{2}\big)}}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow\frac\pi2}\ \frac{{\text{k}\cos{\text{x}}}}{\pi-2\text{x}}=3$
$\Rightarrow\frac{{\text{k}}}{2}\times\lim\limits_{\text{x}-\frac\pi2\rightarrow0}\ \frac{\sin\big(\frac{\pi}{2}-{\text{x}}\big)}{\frac\pi2-{\text{x}}}=3$
$\Rightarrow\frac{\text{k}}{2}\times\lim\limits_{\text{h}\rightarrow0}\ \frac{\sin(-\text{h)}}{-\text{h}}=3$ $\big(\text{Put }\text{x}-\frac\pi2=\text{h}\big)$
$\Rightarrow\frac{\text{k}}{2}\times\lim\limits_{\text{h}\rightarrow0}\ \frac{\sin\text{h}}{\text{h}}=3$ $[\sin(-\theta)=-\sin\theta]$
$\Rightarrow\frac{\text{k}}{2}=3$ $\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big)$
$\Rightarrow\text{k}=6$
Hence, the value of k is 6.
View full question & answer→Question 613 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos7\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos3\text{x}-\cos7\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{-2\sin\big(\frac{3\text{x}+7\text{x}}{2}\big)\sin\big(\frac{3\text{x}-7\text{x}}{2}\big)}{\text{x}^2}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{-2\sin5\text{x}\sin\big(\frac{-4\text{x}}{2}\big)}{\text{x}^2}\Bigg)$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin5\text{x}}{\text{x}}\Big)\times\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin(-2\text{x})}{\text{x}}\Big)$
$=\Big(-2\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{5\text{x}}\Big)\times\Big(-1\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)$
$=(-2\times5)(-1\times2)$
$=20$
View full question & answer→Question 623 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{2}{\text{x}^2-4\text{x}+3}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{2}{\text{x}^2-4\text{x}+3}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{2}{(\text{x}-3)(\text{x}-1)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\Big(\frac{\text{x}-1-2}{(\text{x}-1)(\text{x}-3)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\bigg(\frac{\text{x}-3}{(\text{x}-1)(\text{x}-3)}\bigg)$
$=\lim\limits_{\text{x}\rightarrow3}\frac{1}{\text{x}-1}$
$=\frac{1}{3-1}$
$=\frac12$
View full question & answer→Question 633 Marks
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{x}^2-3\text{x}+2}{\text{x}^3-2\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{x}^2-3\text{x}+2}{\text{x}^3-2\text{x}^2}$$=\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{x}^2-\text{x}-2\text{x}+2}{\text{x}^2(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{x}(\text{x}-1)-2(\text{x}-1)}{\text{x}^2(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow0^-}\frac{(\text{x}-1)(\text{x}-2)}{\text{x}^2(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow0^-}\frac{(\text{x}-1)}{\text{x}^2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(0-\text{h}-1)}{(0-\text{h})^2}$
$\Rightarrow\frac{-\text{h}}{\text{h}^2}=\frac{-1}{\text{h}}=\frac{-1}{0}=-\infty$
View full question & answer→Question 643 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x}+\sin\text{x})^{\frac{1}{\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}(\cos\text{x}+\sin\text{x})^{\frac{1}{\text{x}}}$
$=\ \lim\limits_{\text{x}\rightarrow0}(1+(\cos\text{x}+\sin\text{x}-1))^{\frac{1}{\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{(\cos\text{x}+\sin\text{x}-1)}{\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{(\sin\text{x}-(1-\cos\text{x}))}{\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sin\text{x}-2\sin^2\big(\frac{\text{x}}{2}\big)\big)}{\text{x}}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{x}}{2}\big)\sin\big(\frac{\text{x}}{2}\big)}{2\big(\frac{\text{x}}{2}\big)}}$
$=\text{e}^{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\frac{\text{x}}{2}\big)\sin\big(\frac{\text{x}}{2}\big)}{\big(\frac{\text{x}}{2}\big)}}$
$=\text{e}^{1-0}$
$=\text{e}$
View full question & answer→Question 653 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\sin\text{x}}{\cos\text{x}(1-\cos\text{x})}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\sin\text{x}}{\cos\text{x}\big(2\sin^2\frac{\text{x}}{2}\big)}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)}{\cos\text{x}\big(2\sin^2\frac{\text{x}}{2}\big)}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\cos\frac{\text{x}}{2}}{\cos\text{x}\frac{\sin{\text{x}}{}}{2}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\cos\frac{\text{x}}{2}}{\cos\text{x}\frac{\sin\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1}{\cos\text{x}\times\frac{\frac{\tan\text{x}}{2}}{\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1}{\cos\text{x}}\times\frac{1}{\cos\text{x}\times\frac{\frac{\tan\text{x}}{2}}{\frac{\text{x}}{2}}\times\frac12}$
$=1\times1\times2$
$=2$
View full question & answer→Question 663 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{1-\sin\text{x}}{\big(\frac{\pi}{2}-\text{x}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{1-\sin\text{x}}{\big(\frac{\pi}{2}-\text{x}\big)}$
If $\text{x}\rightarrow\frac{\pi}{2},\frac{\pi}{2}-\text{x}\rightarrow0$
Let $\frac\pi2-\text{x}=\text{y}$ they y → 0
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\sin\big(\frac{\pi}{4}-\text{y}\big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos\text{y}}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2}$
$=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14$ $\Big[\because\lim\limits_{\text{x}\rightarrow{0}}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=2\times1\times\frac14$
$=\frac12$
View full question & answer→Question 673 Marks
Let $a_1, a_2, ..., a_n$ be fixed real numbers such that $f(x) = (x - a_1)(x - a_2) ...(x - a_n)$ What is $\lim\limits_{\text{x}\rightarrow\text{a}_1}\text{f(x)}?$ For $\text{a}\ne\text{a}_1,\text{a}_2,\dots\text{a}_\text{n}$ compute $\lim\limits_{\text{x}\rightarrow{\text{a}}}\text{f(x)}.$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow\text{a}_1}({\text{x}-{\text{a}_1}})(\text{x}-\text{a}_2)\dots(\text{x}-\text{a}_\text{n})$ [Putting limits $x \rightarrow a_1$]
$\Rightarrow(\text{a}_1-\text{a}_1)(\text{a}_1-\text{a}_2)\dots(\text{a}_1-\text{a}_\text{n})$
$\Rightarrow0$
And,
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\text{a}}}(\text{x}-\text{a}_1)(\text{x}-\text{a}_2)\dots(\text{x}-\text{a}_\text{n})$ [Putting limit $x \rightarrow a$]
$\Rightarrow(\text{a}-\text{a}_1)(\text{a}-\text{a}_2)\dots(\text{a}-\text{a}_\text{n}).$
View full question & answer→Question 683 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^2-3}{{\text{x}^2+3\sqrt{3}\text{x}-12}}$
Answer$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^2-3}{{\text{x}^2+3\sqrt{3}\text{x}-12}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)}{\text{x}^2+4\sqrt{3\text{x}}-\sqrt{3\text{x}}-12}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)}{\text{x}\big(\text{x}+4\sqrt{3}\big)-\sqrt{3}\big(\text{x}+4\sqrt{3}\big)}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)}{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+4\sqrt{3}\big)}$
$=\frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}+4\sqrt{3}}=\frac{2\sqrt{3}}{5\sqrt{3}}$
$=\frac25$
View full question & answer→Question 693 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{1-\text{x}^{\frac{-1}{3}}}{1-\text{x}^{\frac{-2}{3}}}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{1-\text{x}^{\frac{-1}{3}}}{1-\text{x}^{\frac{-2}{3}}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{1-\frac{1}{\text{x}^{\frac{1}{3}}}}{{1-\frac{1}{\text{x}^{\frac23}}}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\Big(\text{x}^{\frac13}-1\Big)}{\Big(\text{x}^{\frac{1}{3}}-1\Big)\Big(\text{x}^{\frac13}+1\Big)}\times\text{x}^{\frac{1}{3}}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^{\frac{1}{3}}}{\text{x}^{\frac{1}{3}}+1}$
$=\frac{1}{1+1}$
$=\frac{1}{2}$
View full question & answer→Question 703 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$
Answer$\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}^2+2\sqrt{2}\text{x}-\sqrt{2}\text{x}-4}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}\big(\text{x}+2\sqrt{2}\big)-\sqrt{2\big(\text{x}+2\sqrt{2}\big)}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\big(\text{x}+2\sqrt{2}\big)\big(\text{x}-\sqrt{2}\big)}$
$=\frac{\sqrt{2}+\sqrt{2}}{\sqrt{2}+2\sqrt{2}}=\frac{2\sqrt{2}}{3\sqrt{2}}$
$=\frac23$
View full question & answer→Question 713 Marks
Prove that $\lim\limits_{\text{x}\rightarrow\text{a}^+}\ [\text{x}]=[\text{a}]$ for all $\text{a }\in\text{ R}.$ also prove that $\lim\limits_{\text{x}\rightarrow1^-}\ [\text{x}]=0.$
Answer$\lim\limits_{\text{x}\rightarrow\text{a}^+}\ [\text{x}]$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}[\text{a}+\text{h}]=[\text{a}]$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\ [\text{x}]=[\text{a}]\forall\text{a }\in\text{ R}$
Also,
$\lim\limits_{\text{x}\rightarrow1^-}\ [\text{x}]$
$=\lim\limits_{\text{h}\rightarrow0}\ [1-\text{h}]$
$=0$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^-}\ [\text{x}]=0$
View full question & answer→