Question 15 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{3}}}\frac{\sqrt{3}-\tan\text{x}}{\pi-3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{3}}}\frac{\sqrt{3}-\tan\text{x}}{\pi-3\text{x}}$
If $\text{x}\rightarrow\frac{\pi}{3},\frac{\pi}{3}-\text{x}\rightarrow0,\pi-3\text{x}\rightarrow0$
Let $\frac\pi3-\text{x}=\text{y}$ they y → 0
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{3}-\tan\big(\frac{\pi}{3}-\text{y}\big)}{3\big(\frac{\pi}{3}-\text{x}\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\begin{pmatrix}\frac{\bigg(\sqrt{3}-\frac{\tan\frac\pi3-\tan\text{y}}{1+\tan\frac\pi3.\tan\text{y}}\bigg)}{3\text{y}}\end{pmatrix}$
$=\lim\limits_{\text{y}\rightarrow{0}}\begin{pmatrix}\frac{\Big(\sqrt{3}-\frac{\sqrt{3}-\tan\text{y}}{1+\sqrt{3}\tan\text{y}}\Big)}{3\text{y}}\end{pmatrix}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\big(\sqrt{3}-\tan\text{y}-\sqrt{3}+\tan\text{y}\big)}{3\big(1+\sqrt{3}\tan\text{y}\big)\text{y}}$
$=\frac{4}{3}\times\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\Big(1+\sqrt{3}\frac{\tan\text{y}}{\text{y}}\times\text{y}\Big)}$
$=\frac{4\times1}{3}\times\frac{1}{1+0}$
$=\frac{4}{3}$
View full question & answer→Question 25 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow7}\frac{4-\sqrt{9+\text{x}}}{1-\sqrt{8-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow7}\frac{4-\sqrt{9+\text{x}}}{1-\sqrt{8-\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow7}\frac{\big(4-\sqrt{9+\text{x}}\big)}{1-\sqrt{8-\text{x}}}\times\frac{\big(4+\sqrt{9+\text{x}}\big)}{\big(4+\sqrt{9+\text{x}}\big)}\times\frac{\big(1+\sqrt{8-\text{x}}\big)}{\big(\sqrt{1+\sqrt{8+\text{x}}}\big)}$
$=\lim\limits_{\text{x}\rightarrow7}\frac{\big((4)^2-\big(\sqrt{9+\text{x}}\big)^2\big)}{\big((1)^2-\big(\sqrt{8-\text{x}}\big)^2\big)}\times\frac{1+\sqrt{8-\text{x}}}{4+\sqrt{9+\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow7}\frac{(16-9-\text{x})\times\big(1+\sqrt{8-\text{x}}\big)}{(1-8+\text{x})\times\big(4+\sqrt{9+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow7}\frac{7-\text{x}}{(-7)(7-\text{x})}\frac{\big(1+\sqrt{8-\text{x}}\big)}{\big(4+\sqrt{9+\text{x}}\big)}$
$=\frac{1}{(-1)}\times\frac{(1+1)}{(4+4)}=\frac{-1}{4}$
View full question & answer→Question 35 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{-2\sin\big(\frac{2\text{x}+8\text{x}}{2}\big)\sin\big(\frac{2\text{x}-8\text{x}}{2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin^2\text{x}}{\sin5\text{x}\times\sin(-3\text{x})}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}}{-\big(\lim\limits_{\text{x}\rightarrow0}\sin5\text{x}\big)\big(-\lim\limits_{\text{x}\rightarrow0}\sin3\text{x}\big)}$
$=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{5\text{x}}\times5\text{x}\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\big)\times3\text{x}}$
$=\frac{1\times\text{x}^2}{1\times5\text{x}\times1\times3\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac{\text{x}^2}{15\text{x}^2}$
$=\frac{1}{15}$
View full question & answer→Question 45 Marks
Evaluate the following limit:
If $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-1}{\text{x}-1},$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^4-1}{\text{x}-1}\ \cdots{\text{(i})}$
$\text{L.H.S}=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{3}-\text{a}^3}{\text{x}-\text{a}}$
$=3(\text{a})^{3-1}$
$=3\text{a}^{2}\ \cdots{\text{(ii})}$
$\text{R.H.S}=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^4-1}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow5}\frac{\text{x}^4-1}{\text{x}-1}$
$=4(1)^{4-1}$
$=4\ \cdots{(\text{iii})}$
Substituting (ii) and (iii) in (i),
$3\text{a}^8=4$
$\Rightarrow\text{a}^{2}=\frac43$
$\Rightarrow\text{a}=\pm\frac{2}{\sqrt{3}}$
View full question & answer→Question 55 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$
Answer$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}(\text{x}+1)+\sin(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}}{\text{x}-2}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{(\text{x}-2)}\Bigg(\frac{1}{\text{x}+\frac{\sin(\text{x}+1)}{\text{x}+1}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\text{x}-2}\times\frac{1}{\lim\limits_{\text{x}\rightarrow-1}(\text{x})+\lim\limits_{\text{x}\rightarrow-0}\sin\frac{\text{x}+1}{\text{x}+1}}$
$=\Big(\frac{1}{-1-2}\Big)\times\frac{1}{(-1)+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{1}{0}$ $\Big[\because\frac10=\infty\Big]$
$=\infty$
View full question & answer→Question 65 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}\sqrt{\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}\sqrt{\text{x}+2}$
$=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big[\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big]\times\frac{\big[\sqrt{\text{x}+1}+\sqrt{\text{x}}\big]}{\big[\sqrt{\text{x}+1}+\sqrt{\text{x}}\big]}\times\frac{\sqrt{\text{x}+2}\times\sqrt{\text{x}+2}}{\sqrt{\text{x}+2}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{(\text{x}+1-\text{x})}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\times\frac{(\text{x}+2)}{\sqrt{\text{x}+2}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1(\text{x}+2)}{\text{x}\big(\sqrt{\text{x}+1}+\sqrt{{\text{x}}}\big)\big(\sqrt{\text{x}+2}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\text{x}\Big(1+\frac{2}{\text{x}}\Big)}{\sqrt{\text{x}}\bigg(\sqrt{1+\frac{1}{\text{x}}}+1\bigg)\bigg(\sqrt{1+\frac{2}{\text{x}}}\bigg)\sqrt{\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\Big(1+\frac{2}{\text{x}}\Big)}{\bigg(\sqrt{1+\frac{1}{\text{x}}}+\sqrt{1}\bigg)\bigg(\sqrt{1+\frac{2}{\text{x}}}\bigg){}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{(1+0)}{(1+1)\times1}=\frac12$
View full question & answer→Question 75 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\text{n}\sin\Big(\frac{\pi}{4\text{n}}\Big)\cos\Big(\frac{\pi}{4\text{n}}\Big)$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\text{n}\sin\Big(\frac{\pi}{4\text{n}}\Big)\cos\Big(\frac{\pi}{4\text{n}}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(\text{n}\sin\frac{\pi}{4\text{n}}\cos\frac{\pi}{4\text{n}}\Big)\times\frac12$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{n}\times\sin\frac{\pi}{2\text{n}}\times\frac12$
$\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}\rightarrow0, $ let $\frac{1}{\text{n}}=\text{y}$
$=\frac12\lim\limits_{\frac{1}{\text{n}}\rightarrow\infty}\frac{1}{\text{y}}\sin\Big(\frac{\pi}{2}\Big)\Big(\frac{1}{\text{n}}\Big)$
$=\frac12\lim\limits_{{\text{y}}\rightarrow\infty}\frac{\sin\big(\frac\pi2\big)\text{y}}{\text{y}}$
$=\frac12\Bigg(\lim\limits_{{\text{y}}\rightarrow\infty}\frac{\sin\big(\frac{\pi\text{y}}2\big)}{\frac{\pi\text{y}}{2}}\Bigg)\times\frac\pi2$
$=\frac12\times1\times\frac\pi2$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac\pi4$
View full question & answer→Question 85 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+3\Big)\Big(\text{x}^{\frac{1}{3}}-3\Big)}{\text{x}-27}$
Answer$\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+3\Big)\Big(\text{x}^{\frac{1}{3}}-3\Big)}{\text{x}-27}$
$=\lim\limits_{\text{x}\rightarrow{27}}\frac{\Big(\text{x}^\frac{1}{3}+9\Big)}{\text{x}-27}$
$=\lim\limits_{\text{x}\rightarrow{27}}\frac{\text{x}^\frac{2}{3}-27^{\frac{2}{3}}}{\text{x}-27}$
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
$=\frac23(27)^{\frac23-1}$
$=\frac23(27)^{\frac{-1}{3}}$
$=\frac23\times\frac{1}{(27)^{\frac{1}{3}}}$
$=\frac{2}{3}\times\frac{1}{3}$
$=\frac29$
View full question & answer→Question 95 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 105 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\cot^2\text{x}-3}{\text{cosec x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\cot^2\text{x}-3}{\text{cosec x}-2}$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec}^2\text{x}-1\big)-3}{\text{cosec x}-2}$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec}^2\text{x}-4\big)}{\text{cosec x}-2}$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\frac{\big(\text{cosec }\text{x}-2\big)\big(\text{cosec }\text{x}+2\big)}{\text{cosec x}-2}$
$=\lim\limits_{\text{x}\rightarrow{\frac{\pi}{6}}}\text{cosec }\text{x}+2$
$=\text{cosec}\frac{\pi}{6}+2$
$=2+2$
$=4$
View full question & answer→Question 115 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big\{\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big\}$
$=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\Big[\sqrt{\text{x}^2+1}-\sqrt{\text{x}^2-1}\Big]\times\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}\big)}{\big(\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2+1}\big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\text{x}\frac{\text{x}\big(\text{x}^2+1-\text{x}^2+1\big)}{\sqrt{\text{x}^2+1}+\sqrt{\text{x}^2-1}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}(2)}{\text{x}\Big(\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{2}{\sqrt{1+\frac{1}{\text{x}^2}}+\sqrt{1-\frac{1}{\text{x}^2}}}$
$=\frac{2}{2}=1$
$=1$
View full question & answer→Question 125 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}}$
$\Rightarrow\text{x}\rightarrow{\pi},\text{x}-{\pi}\rightarrow0,$ let $\text{y}=\text{x}-\pi$
$\Rightarrow\lim\limits_{\text{x}-{\pi}\rightarrow{{0}}}\frac{1+\cos\text{x}}{\tan^2\text{x}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos\text{y}}{\tan^2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$
$=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}\times\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan\text{y}}{\text{y}}\Big)^2\times\text{y}^2}$
$=2\times1\times\frac{\text{y}^2}{4}\times\frac{1}{1\times\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$
$=\frac{1}{2}$
View full question & answer→Question 135 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\big(\sqrt{1+\text{x}}-\sqrt{1-\text{x}}\big)}\times\frac{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}\big)^2-\big(\sqrt{1-\text{x}}\big)^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{1+\text{x}-1+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{2\text{x}}$
$=\frac12\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{\text{x}}\Big)\text{x}$
$=\frac{1}{2}\lim\limits_{\text{x}\rightarrow0}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$
$=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$
$=\frac{1}{2}(1+1)=\frac22$
$=1$
View full question & answer→Question 145 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$
Answer$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sqrt{5+\cos\text{x}-2}}{(\pi-\text{x})^2}$
$⇒ \text{x} → \pi,$ then $\pi-\text{x}\rightarrow0,$ let $\pi-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5+\cos(\pi-\text{y})}-2}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{5-\cos\text{y}}-2}{\text{y}^2}\times\frac{\big(\sqrt{5-\cos\text{y}}+2\big)}{\big(\sqrt{5-\cos\text{y}}+2\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{(5-\cos\text{y}-4)}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{5-\cos\text{y}}+2\big)}$
$=2\times\frac14\times\frac{1}{\sqrt{4}+2}=2\times\frac14\times\frac14$
$=\frac18$
View full question & answer→Question 155 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\sqrt{\text{x}}-\sqrt{2}}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)\big(\sqrt{\text{x}}+\sqrt{2}\big)}{\big(\sqrt{\text{x}}-\sqrt{2}\big)\big(\sqrt{\text{x}}+\sqrt{2}\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)\big(\sqrt{\text{x}}+\sqrt{2}\big)}{(\text{x}-2)}$
$=\sqrt{2}+\sqrt{2}$
$=2\sqrt{2}$
View full question & answer→Question 165 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\tan\text{x}}{2\sin^2\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\frac{\tan\text{x}}{\text{x}}}{\frac{2\sin^2\text{x}}{\text{x}^2}}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}}{2\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$
$=\frac{1}{2\times1}$
$=\frac{1}{2}$
View full question & answer→Question 175 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{ax}-\cos\text{bx}}{\cos\text{cx}-1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1-2\sin^2\big(\frac{\text{ax}}{2}\big)-1+2\sin^2\big(\frac{\text{bx}}{2}\big)}{1-2\sin^2\big(\frac{\text{cx}}{2}\big)-1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\sin^2\big(\frac{\text{ax}}{2}\big)+2\sin^2\big(\frac{\text{bx}}{2}\big)}{-2\sin^2\big(\frac{\text{cx}}{2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-\sin^2\big(\frac{\text{ax}}{2\text{ax}}\big)4\text{a}^2\text{x}^2+\sin\big(\frac{\text{bx}}{2}\big)4\text{b}^2\text{x}^2}{-\sin^2\big(\frac{\text{cx}}{2}\big)4\text{c}^2\text{x}^2}$
$=\frac{-\text{a}^2+\text{b}^2}{-\text{c}^2}$
$=\frac{\text{a}^2-\text{b}^2}{\text{c}^2}$
View full question & answer→Question 185 Marks
Evaluate the following limit:
If $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x}),$ find all possible value of a.
AnswerIf $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{9}-\text{a}^9}{\text{x}-\text{a}}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x})\ \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}\ \cdots{\text{(ii})}$
$\text{R.H.S}=\lim\limits_{\text{x}\rightarrow5}(4+\text{x})$
$=4+5=9\ \cdots{\text{(iii})}$
Substituting (ii) and (iii) in (i)
$9\text{a}^8=9$
$\Rightarrow\text{a}^{8}=1$
$\Rightarrow\text{a}^4=1$
$\text{a}^2=1$
$\Rightarrow\text{a} = 1\text{ and a} = -1$
View full question & answer→Question 195 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin2\pi\text{ x}}$
⇒ x → 1, then x - 1 →0, let x - 1 = y
$=\lim\limits_{(\text{x}-1)\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin2\pi\text{ x}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{-\text{y}(1+\text{y}+1)}{\sin2\pi(\text{y}+1)}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin(2\pi\text {y}+2\pi)}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin2\pi\text {y}}$
$=-\lim\limits_{\text{y}\rightarrow{0}}(\text{y}+2)\times\frac{\text{y}}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\sin\frac{2\pi\text {y}}{\text {y}\times2\pi}\Big)\times2\pi\text {y}}$
$=-2\times\frac{1}{1\times2\pi}$
$=-\frac{1}{\pi}$
View full question & answer→Question 205 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}\big)\times\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}+\text{x}^2\big)-(\text{x}+1)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$
$=\frac{1}{2\big(\sqrt{1}+\sqrt{1}\big)}$
$=\frac{1}{2\times2}$
$=\frac14$
View full question & answer→Question 215 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$
$\Rightarrow\text{x}\rightarrow\frac{\pi}{4},\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac{\pi}{4}=\text{y}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{x}}=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\Big(1-\sin2\big(\text{y}+\frac{\pi}{4}\big)\Big)}{1+\cos4\big(\text{y}+\frac{\pi}{4}\big)}$
$=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\Bigg(\frac{1-\sin\big(\frac\pi2+2\text{y}\big)}{1+\cos(\pi+4\text{y})}\Bigg)$
$=\lim\limits_{\text{x}-\frac\pi4\rightarrow{0}}\frac{1-\cos2\text{y}}{1-\cos4\text{y}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\text{y}}{2\sin^22\text{y}}$
$=\frac{\lim\limits_{\text{y}\rightarrow{0}}\sin^2\text{y}}{\lim\limits_{\text{y}\rightarrow{0}}\sin^22\text{y}}$
$=\frac{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}\Big)^2\times\text{y}^2}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\text{y}^2}$
$=\frac{1\times\text{y}^2}{1\times4\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac14$
View full question & answer→Question 225 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$
$=\frac{2\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}+\lim\limits_{\text{x}\rightarrow0}\tan^2\text{x}}{\lim\limits_{\text{x}\rightarrow0}\text{x}\sin\text{x}}$
$=\frac{\Big(2\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2\Big)+\big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)\times\text{x}^2}$
$=\frac{\big(2\times1\times\text{x}^2\big)+\big(1\times\text{x}^2\big)}{\big(1\times\text{x}^2\big)}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{3\text{x}^2}{\text{x}^2}$
$=3$
View full question & answer→Question 235 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}$
$\text{x}\rightarrow{\pi},$ then $\text{x}-{\pi}\rightarrow0,$ let $\text{x}-\pi=\text{y}$
$\Rightarrow\lim\limits_{{\text{x}\rightarrow{\pi}}}\frac{\big(\sqrt{2+\cos\text{x}}-1\big)}{(\pi-\text{x})^2}=\lim\limits_{{\text{x}-{\pi}}\rightarrow0}\frac{\sqrt{2+\cos(\text{x})}-1}{(-1)^2(\text{x}-\pi)^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2+\cos(\pi+\text{y})}-1}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\sqrt{2-\cos\text{y}}-1}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\big(\sqrt{2-\cos\text{y}}-1\big)\big(\sqrt{2-\cos\text{y}}+1\big)}{\text{y}^2\big(\sqrt{2\cos\text{y}}+1\big)}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{(2\cos\text{y}-1)}{\big(\sqrt{2-\cos\text{y}+1}\big)\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{(1-\cos\text{y})}{\big(\sqrt{2-\cos\text{y}}+1\big)\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\text{y}^2\big(\sqrt{2-\cos\text{y}}+1\big)}$
$=2\lim\limits_{\text{y}\rightarrow0}\Bigg(\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac14\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\sqrt{2-\cos0+1}}$
$=2\times\frac14\times\frac{1}{\sqrt{2}-1+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{1}{4}$
View full question & answer→Question 245 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 255 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{ x}-3}{\text{cosec}\text{ x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{x}-3}{\text{cosec }\text{x}-2}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{\text{cosec}^2\text{x}-4}{\text{cosec }\text{x}-2}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}\frac{(\text{cosec }\text{x}-2)(\text{cosec x}+2)}{(\text{cosec }\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{6}}(\text{cosec }\text{x}+2)$
$=\text{cosec}\frac{\pi}{6}+2$
$=2+2=4$
View full question & answer→Question 265 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+\ \dots+\text{n}^3}{(\text{n}-1)^4}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^3+2^3+3^3\ \dots+\text{n}^3}{(\text{n}-1)^4}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big[\frac12(\text{n})(\text{n}+1)\Big]^2}{(\text{n}-1)^4}$ $\bigg[1^3+2^3+3^3+\ \cdots+\text{n}^3=\Big(\frac12\text{n}(\text{n}+1)\Big)^2\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\frac{\frac14\text{n}^2\big(\text{n}^2+1+2\text{n}\big)}{(\text{n}-1)^4}\Bigg)$
$=\frac14\lim\limits_{\text{n}\rightarrow\infty}\bigg(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{(\text{n}-1)^2(\text{n}-1)^2}\bigg)$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\Big(\frac{\text{n}^4+\text{n}^2+2\text{n}^3}{\text{n}^4+\text{n}^2-2\text{n}^3+\text{n}^2+1-2\text{n}-2\text{n}^3-2\text{n}+4\text{n}^2}\Big)$
$=\frac{1}{4}\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}\Big)}{\Big(1+\frac{1}{\text{n}^2}+\frac{2}{\text{n}}+\frac{1}{\text{n}^2}+\frac{1}{\text{n}^4}-\frac{2}{\text{n}^3}-\frac{2}{\text{n}}-\frac{2}{\text{n}^3}+\frac{4}{\text{n}^2}\Big)}$
$=\frac{1}{4}\Big(\frac14\Big)$
$=\frac14$
View full question & answer→Question 275 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{cosec x}-\cot\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{cosec x}-\cot\text{x}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big)\times\frac{1}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}\Big(\frac{1-\cos\text{x}}{\text{x}}\Big)\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{1}{\sin\text{x}}\bigg(\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}}\bigg)\bigg)$
$=2\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{1}{\frac{\sin\text{x}}{\text{x}}}\times\text{x}\bigg(\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)\times\frac{\text{x}}{4}\Bigg)$
$=2\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{1}{\frac{\sin\text{x}}{\text{x}}}\bigg)\times\frac{1}{\text{x}}\times\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\bigg)\times\frac{\text{x}}{4}$
$=2\times\frac{1}{\text{x}}\times\frac{\text{x}}{4}$
$=\frac{1}{2}$
View full question & answer→Question 285 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 295 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→Question 305 Marks
Find $\lim\limits_{\text{x}\rightarrow-\frac{5}{2}}{[\text{x}]}.$
Answer$\lim\limits_{\text{x}\rightarrow-\frac{5}{2}}{[\text{x}]}$
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}}{[\text{x}]}=\Big[\frac{5}{2}\Big],$
$=[2.5]=2$ [By definition of geatest integer function]
$\Rightarrow\ \lim\limits_{\text{x}\rightarrow\frac{5}{2}}{[\text{x}]}=2$
View full question & answer→Question 315 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$
$=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$
If t = x - a
Then, as x → a, t → 0
$\therefore\ \lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{(\text{a}\sin\text{x}-\text{x}\sin\text{a})}{\text{ax}(\text{x}-\text{a})}$
$=\lim\limits_{\text{t}\rightarrow{0}}\frac{\big(\text{a}\sin(\text{t}+\text{a})-(\text{t}+\text{a})\sin\text{a}\big)}{\text{a}(\text{t}+\text{a})\text{t}}$
$=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}\cot\text{t}-\text{t}\sin\text{a}-\text{a}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{a}}$
$=\lim\limits_{\text{t}\rightarrow{0}}\frac{\text{a}\sin\text{t}\cos\text{a}+\text{a}\sin\text{a}(\cos\text{t}-1)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$
$=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\cos\text{a}+\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)-\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$
$=\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{t}\cos\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}+\lim\limits_{\text{t}\rightarrow0}\frac{\text{a}\sin\text{a}\Big(2\sin^2\big(\frac{\text{t}}{2}\big)\Big)}{\text{a}(\text{t}+\text{a})\text{t}}-\lim\limits_{\text{t}\rightarrow0}\frac{\text{t}\sin\text{a}}{\text{a}(\text{t}+\text{a})\text{t}}$
$=\frac{\text{a}\cos\text{a}}{\text{a}^2}+0-\frac{\sin\text{a}}{\text{a}^2}$
$=\frac{\text{a}\cos\text{a}-\sin\text{a}}{\text{a}^2}$
View full question & answer→Question 325 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{3+\text{x}}-\sqrt{5-\text{x}}}{\text{x}^2-1}$$=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\sqrt{3+\text{x}}-\sqrt{5-\text{x}}\big)}{\big(\text{x}^2-1\big)}\times\frac{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}{\big(\sqrt{3+\text{x}}+\sqrt{5-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(3+\text{x})-(5-\text{x})}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{-2+2\text{x}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{2}{(\text{x}+1)\big(\sqrt{3+\text{x}}+\sqrt{5+\text{x}}\big)}$
$=\frac{2}{(1+1)\big(\sqrt{3+1}+\sqrt{5-1}\big)}=\frac{2}{(2)(2+2)}$
$=\frac14$
View full question & answer→Question 335 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}^\circ-\sin2\text{x}^\circ}{\text{x}^3}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}^\circ-\sin2\text{x}^\circ}{\text{x}^3}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}-\sin\frac{2\pi\text{x}}{180}}{\text{x}^3}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}-2\sin\frac{\pi\text{x}}{180}\cos\frac{\pi\text{x}}{180}}{\text{x}^3}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\frac{\pi\text{x}}{180}\big(2\sin^2\frac{\pi\text{x}}{360}\big)}{\text{x}^3}$
$=4\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\text{x}}\bigg)\times\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\text{x}}\bigg)\times\bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\text{x}}\bigg)$
$=4\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\frac{\pi\text{x}}{180}}\times\frac{\pi}{180}\Bigg)\times\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\frac{\pi\text{x}}{360}}\times\frac{\pi}{360}\Bigg)\times\Bigg(\lim\limits_{\text{x} \rightarrow0}\frac{\sin\frac{\pi\text{x}}{360}}{\frac{\pi\text{x}}{360}}\times\frac{\pi}{360}\Bigg)$
$=4\times\frac{\pi}{180}\times\frac{\pi}{360}\times\frac{\pi}{360}$
$=\Big(\frac{\pi}{180}\Big)^3$
View full question & answer→Question 345 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+\text{x}-2\big]}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+2\text{x}-\text{x}-2\big]}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)(\text{x}-1)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)}{3(\text{x}+2)\big(\sqrt{\text{x}}+1\big)}$
$=\frac{(2-3)}{3(1+2)(1+1)}$
$=\frac{-1}{3\times3\times2}$
$=-\frac{1}{18}$
View full question & answer→Question 355 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{\text{x}^2+1}-\sqrt{5}}{\text{x}-2}\times\frac{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{\big({\text{x}^2+1}-{5}\big)}{(\text{x}-2)\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)(\text{x}-2)}{(\text{x}-1)\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)}{\big(\sqrt{\text{x}^2+1}+\sqrt{5}\big)}$
$=\frac{(2+2)}{\sqrt{4+1}+\sqrt{5}}$
$=\frac{4}{2\sqrt{5}}=\frac{2}{\sqrt{5}}$
View full question & answer→Question 365 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 375 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 385 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 395 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^3+\text{x}^2+4\text{x}+12}{\text{x}^3+3\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^3+\text{x}^2+4\text{x}+12}{\text{x}^3+3\text{x}+2}$
$=\lim\limits_{\text{x}\rightarrow-2}\frac{(\text{x}+2)\big(\text{x}^2+\text{x}+6\big)}{(\text{x}+2)\big(\text{x}^2-2\text{x}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow-2}\frac{\text{x}^2+\text{x}+6}{\text{x}^2-2\text{x}+1}$
$=\frac{(-2)^2-(-2)+6}{(-2)^2-2(-2)+1}$
$=\frac{4+2+6}{4+4+1}$
$=\frac{12}{9}$
$=\frac{4}{3}$
View full question & answer→Question 405 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^{3}+1}{\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^{3}+1}{\text{x}+1}$$=\lim\limits_{\text{x}\rightarrow{-1}}\frac{\text{x}^3-(-1)^3}{\text{x}-(-1)}$ [Dividing numerator and denominator by x - 1]
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$
Here, n = 3, a = -1
$\Rightarrow\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3-(-1)^3}{\text{x}-(-1)}=\text{na}^{\text{n}-1}$
$=3(-1)^{3-1}$
$=3(-1)^2$
$=3$
View full question & answer→Question 415 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}}{\text{x}\sqrt{\text{a}^2+\text{ax}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}}{\text{x}\sqrt{\text{a}^2+\text{ax}}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{\text{a}+\text{x}}-\sqrt{\text{a}}\big)}{\text{x}\sqrt{\text{a}^2+\text{ax}}}\times\frac{\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}{\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a}+\text{x})-\text{a}}{\text{x}\sqrt{\text{a}^2+\text{ax}}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{x}\sqrt{\text{a}^2+\text{ax}}\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{\text{a}^2+\text{ax}}\big)\big(\sqrt{\text{a}+\text{x}}+\sqrt{\text{a}}\big)}$
$=\frac{1}{\text{a}\big(\sqrt{2\text{a}}\big)}$
$=\frac{1}{2\text{a}\sqrt{\text{a}}}$
View full question & answer→Question 425 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}+7\text{x}}{4\text{x}+\sin2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}+7\text{x}}{4\text{x}+\sin2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\frac{\sin3\text{x}}{\text{x}}+\frac{7\text{x}}{\text{x}}\big)}{\big(\frac{4\text{x}}{\text{x}}+\frac{\sin2\text{x}}{\text{x}}\big)}$
$=\frac{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3\Big)+7}{4+\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\Big)\times2}$
$=\frac{3+7}{4+2}$
$=\frac{10}{6}$
$=\frac53$
View full question & answer→Question 435 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{\sin\big(\frac{\text{a}}{2^{\text{n}}}\big)}{\sin\big(\frac{\text{b}}{2^{\text{n}}}\big)}$
$\text{n}\rightarrow\infty,$ then $\frac{1}{\text{n}}=\text{h} \rightarrow0$
$=\frac{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\Bigg)}{\lim\limits_{\text{h}\rightarrow\infty}\sin\Bigg(\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\Bigg)}$
$=\frac{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}{\begin{pmatrix}\lim\limits_{\text{h}\rightarrow\infty}\frac{\sin\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}{\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}\end{pmatrix}}$
$=\frac{1\times\frac{\text{a}}{2^{\frac{1}{\text{h}}}}}{1\times\frac{\text{b}}{2^{\frac{1}{\text{h}}}}}=\frac{\text{a}}{\text{b}}$
View full question & answer→Question 445 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos2\text{x}}{1-\cos\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^2\text{x}}{2\sin^2\frac{\text{x}}{2}}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}(\sin\text{x})^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}\big(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\big)^2}{\lim\limits_{\text{x} \rightarrow0}\big(\sin\frac{\text{x}}{2}\big)^2}$
$=4\lim\limits_{\text{x} \rightarrow0}\cos^2\frac{\text{x}}{2}$
$=4\times1$
$=4$
View full question & answer→Question 455 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$
Answer$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}$
$\Rightarrow\text{x}\rightarrow1,\text{x}-1\rightarrow0,$ let $\text{x}-1=\text{y}\Rightarrow\text{y}\rightarrow0$
$=\lim\limits_{\text{x}\rightarrow{0}}\frac{1-\text{x}^2}{\sin\pi\text{ x}}=\lim\limits_{\text{x}-1\rightarrow{0}}\frac{(1-\text{x})(1+\text{x})}{\sin\pi\text{ x}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{(-\text{y})(1+\text{y}+1)}{\sin\pi(\text{y}+1)}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\sin\pi(\text{y}+1)}$
$=-\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{-\sin\pi(\text{y}+1)}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\text{y}(\text{y}+2)}{\frac{\sin\pi\text{y}}{2}}$
$=\frac{\lim\limits_{\text{y}\rightarrow{0}}\text{y}(\text{y}+2)}{\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\pi\text{y}}{\pi\text{y}}\times\pi\text{y}}$
$=\frac{2}{\pi}$
View full question & answer→Question 465 Marks
Evaluate the following limit:
Evaluate: $\lim\limits_{\text{n}\rightarrow\infty}\frac{1.2+2.3+3.4+\ \cdots+\text{n}(\text{n}+1)}{\text{n}^3}$
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{1.2+2.3+3.4+\ \cdots+\text{n}(\text{n}+1)}{\text{n}^3}$
$=\ \lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}}{\text{n}^3}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{n}(\text{n}+1)\Big[\frac{(2\text{n}+1)+3}{6}\Big]}{\text{n}^3}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\frac{\text{n}(\text{n}+1)(2\text{n}+4)}{6}}{\text{n}^3}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{\Big(1+\frac{1}{\text{n}}\Big)\Big(2+\frac{4}{\text{n}}\Big)}{6}$
$=\frac{1\times2}{6}$
$=\frac{1}{3}$
View full question & answer→Question 475 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$As $\text{x}\rightarrow\frac{\pi}{4},\frac\pi4-\text{x}\rightarrow0,$ let $\frac{\pi}{4}-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\cos\big(\frac\pi4-\text{y}\big)-\sin\big(\frac\pi4-\text{y}\big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\big[\cos\frac\pi4\cos\text{y}+\sin\frac{\pi}{4}\sin\text{y}+\sin\frac\pi4\cos\text{y}-\cos\frac\pi4\sin\text{y}\big]}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\Big(\frac{\cos\text{y}}{\sqrt{2}}+\frac{\sin\text{y}}{\sqrt{2}}+\frac{\cos\text{y}}{\sqrt{2}}-\frac{\sin\text{y}}{\sqrt{2}}\Big)}{\text{y}^2}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{2\cos\text{y}}{\sqrt{2}}}{\text{y}^2}=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sqrt{2}-\sqrt{2}\cos\text{y}}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{(1-\cos\text{y})}{\text{y}^2}$
$=\sqrt{2}\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\frac{\text{y}^2}{4}}\times\frac14$
$=\sqrt{2}\times2\times\frac14\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\bigg)^2$
$=\sqrt{2}\times2\times\frac14\times1$
$=\frac{1}{\sqrt{2}}$
View full question & answer→Question 485 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$$=\lim\limits_{\text{x} \rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}-\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\text{x}^2}{3\text{x}^2}$
$=\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2-\frac{2}{3}\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}$
$=1-\frac{2}{3}\times1$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=1-\frac23$
$=\frac{3-2}{3}$
$=\frac{1}{3}$
View full question & answer→Question 495 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}}{\text{x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}}{\text{x}-2}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\sqrt{1+4\text{x}}-\sqrt{5+2\text{x}}\big)}{(\text{x}-2)}\times\frac{\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}{\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(1+4\text{x})-(5+2\text{x})}{(\text{x}-2)\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{-4+2\text{x}}{(\text{x}-2)\big(\sqrt{1+4\text{x}}+\sqrt{5+2\text{x}}\big)}$
$=\frac{2}{\sqrt{1+8}+\sqrt{5+4}}=\frac{2}{\sqrt{9}+\sqrt{9}}$
$=\frac{2}{6}=\frac13$
View full question & answer→Question 505 Marks
Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{x}^4+7\text{x}^3+46\text{x}+\text{a}}{\text{x}^4+6},$ where a is a non-zero real number.
Answer$\lim\limits_{\text{n}\rightarrow\infty}\frac{\text{x}^4+7\text{x}^3+46\text{x}+\text{a}}{\text{x}^4+6}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1+\frac{7}{\text{x}}+\frac{46}{\text{x}^3}+\frac{\text{a}}{\text{x}^4}}{1+\frac{6}{\text{x}^4}}$
$=1$
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