Question 12 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan^23\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan^23\text{x}}{\text{x}^2}$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{\text{x}}\Big)^2$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\Big)^2$
$=1\times9$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=9$
View full question & answer→Question 22 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\sin\text{kx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\sin\text{kx}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\text{kx}\frac{\sin\text{kx}}{\text{kx}}}$
$=\frac{1}{\text{x}}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\big(\text{a}^\text{mx}-\text{b}^\text{nx}\big)}{\text{x}}}{\frac{\sin\text{kx}}{\text{kx}}}$
$=\frac{1}{\text{x}}\text{log}\frac{\text{a}^\text{m}}{\text{b}^\text{n}}$
View full question & answer→Question 32 Marks
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}}{\sqrt{4\text{x}^2+1}-1}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}}{\sqrt{4\text{x}^2+1}-1}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1}{\sqrt{4+\frac{1}{\text{x}^2}}-\frac{1}{\text{x}}}$
$=\frac{1}{\sqrt{4}-0}$
$=\frac12$
View full question & answer→Question 42 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\sqrt{1-\cos\text{x}}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\sqrt{1-\cos\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{e}^\text{x}-1\big)\big(\sqrt{1+\cos\text{x}}\big)}{\big(\sqrt{1-\cos\text{x}}\big)\big(\sqrt{1+\cos\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{e}^\text{x}-1\big)\big(\sqrt{1+\cos\text{x}}\big)}{\sin\text{x}}$
Both numerator and denominator are both zeros for x = 0 hence limit can not exist.
View full question & answer→Question 52 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-\text{x}-1}{2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-\text{x}-1}{2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{2}-1$
$=1-1$
$=0$
View full question & answer→Question 62 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{log(2+x)}+\text{log0.5}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{log(2+x)}+\text{log0.5}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}\Big(1+\frac{\text{x}}{2}\Big)}{2\Big(\frac{\text{x}}{2}\Big)}$
$=\frac{1}{2}$
View full question & answer→Question 72 Marks
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow0^-}(1+\text{cosec }\text{x})$
Answer$\lim\limits_{\text{x}\rightarrow0^-}(1+\text{cosec }\text{x})$
$=\lim\limits_{\text{x}\rightarrow0^-}1+\text{cosec }(0-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\ 1-\text{cosec}\text{ h}$
$=\lim\limits_{\text{h}\rightarrow0}1-\frac{1}{\sin\text{h}}$
$\Rightarrow1-\frac{1}{0}=-\infty$
View full question & answer→Question 82 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}+2\text{x}}{3\text{x}+2\tan3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}+2\text{x}}{3\text{x}+2\tan3\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{3\frac{\sin\text{x}}{\text{x}}+2}{3+2\tan\frac{3\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}2}{\lim\limits_{\text{x}\rightarrow0}3+\lim\limits_{\text{x}\rightarrow0}\frac{2\tan3\text{x}}{\text{x}}}$
$=\frac{\Big(3\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)+2}{3+\Big(2\times\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\times3\Big)}$
$=\frac{(3\times2)+2}{3+(2\times3)}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ also }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{8}{9}$
View full question & answer→Question 92 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x+2}-\text{e}^2}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x+2}-\text{e}^2}{\text{x}}$
$=\text{e}^2\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}$
$=\text{e}^2$
View full question & answer→Question 102 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\infty}\Big(\text{a}^{\frac{1}{\text{x}}}-1\Big)\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\Big(\text{a}^{\frac{1}{\text{x}}}-1\Big)\text{x}$
Let $ \frac{1}{\text{x}}=\text{h}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{a}^\text{h}-1\big)}{\text{h}}$
$=\text{log a}$
View full question & answer→Question 112 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{4}{\text{x}^3-2\text{x}^2}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{4}{\text{x}^3-2\text{x}^2}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{4}{\text{x}^2(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}^2-4}{\text{x}^2(\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}^2(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)}{\text{x}^2}$
$=\frac{(2+2)}{2^2}=\frac{4}{4}$
$=1$
View full question & answer→Question 122 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sin2\text{x}}{\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sin2\text{x}}{\cos\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{2\sin\text{x}\cos\text{x}}{\cos\text{x}}$ $=2\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\sin\text{x}$ $=2\times\sin\frac{\pi}{2}$ $=2\times1$ $=2$
View full question & answer→Question 132 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3\text{x}^2-4\text{x}^2+6\text{x}-1}{2\text{x}^3+\text{x}^2-5\text{x}+7}$
Answer$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3\text{x}^2-4\text{x}^2+6\text{x}-1}{2\text{x}^3+\text{x}^2-5\text{x}+7}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3-\frac{4}{\text{x}}+\frac{6}{\text{x}^2}-\frac{1}{\text{x}^3}}{2+\frac{1}{\text{x}}-\frac{5}{\text{x}^2}+\frac{7}{\text{x}^3}}$
$=\frac{3-0+0-0}{2+0-0+0}$
$=\frac32$
View full question & answer→Question 142 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{a}^{-\text{x}}-2}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{a}^{-\text{x}}-2}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{2x}-\text{2a}^\text{x}-2}{\text{a}^2.\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{a}^\text{x}-1}{\text{x}}\Big)^2\times\frac{1}{\text{a}^\text{x}}$
$=(\text{log}_e\text{a})^2\times\frac{1}{\text{a}^0}$
$=(\text{log}_e\text{a})^2$
View full question & answer→Question 152 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{3\text{x}}-\text{e}^{2\text{x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{3\text{x}}-\text{e}^{2\text{x}}}{\text{x}}$
$=3\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{3\text{x}}-1\text{}}{3\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{2\text{x}}-1\text{}}{2\text{x}}$
$=3-2$
$=1$
View full question & answer→Question 162 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{3\tan^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{3\tan^2\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^2\text{x}}{3\tan^2\text{x}}$
$=\frac23\lim\limits_{\text{x} \rightarrow0}\frac{\sin^2\text{x}}{\Big(\frac{\sin^2\text{x}}{\cos^2\text{x}}\Big)}$
$=\frac23\lim\limits_{\text{x} \rightarrow0}\cos^2\text{x}$
$=\frac23$
View full question & answer→Question 172 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\text{e}^{-\cos\text{x}}-1}{-\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\text{e}^{-\cos\text{x}}-1}{-\cos\text{x}}$
Let $\text{ x}-\frac{\pi}{2}=\text{h}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-\sin\text{h}}-1}{-\sin\text{h}}$
$=\lim\limits_{\sin\text{h}\rightarrow0}\frac{\text{e}^{-\sin\text{h}}-1}{-\sin\text{h}}$
$=1$
View full question & answer→Question 182 Marks
Evaluate the following one sided limits:
$\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}-3}{\text{x}^2-4}.$
Answer$\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}-3}{\text{x}^2-4}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2+\text{h})-3}{(2+\text{h)}^2-2^2}$ $\Big[\because\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h)}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2-3+\text{h})}{(2+\text{h}-2)(2+\text{h}+2)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{h}-1)}{(\text{h})(4+\text{h})}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\frac{1}{\text{h}}}{4+\text{h}}$
$\frac{1-\frac{1}{0}}{4}=-\infty$
View full question & answer→Question 192 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{1-\cos6\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{1-\cos6\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{2\sin^23\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$ $\big(1-\cos2\theta=2\sin^2\theta\big)$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{2}\sin3\text{x}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sin3\text{x}}{\big(\frac{\pi}{3}-\text{x}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin3\big(\frac{\pi}{3}+\text{h}\big)}{\frac{\pi}{3}-\big(\frac{\pi}{3}-\text{x}\big)}$ $\big(\text{Put x}=\frac{\pi}{3}+\text{h}\big)$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(\pi+3\text{h})}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\sin3\text{h}}{-\text{h}}$ $[\sin(\pi+\theta)=-\sin\theta]$
$=3\times\lim\limits_{\text{h}\rightarrow0}\frac{\sin3\text{h}}{3\text{h}}$
$3\times1$ $\Big(\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big)$
$=3$
View full question & answer→Question 202 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}$
$=\text{log e }\times 1$
$=1$
View full question & answer→Question 212 Marks
Find:
$\lim\limits_{\text{x}\rightarrow1}\ [\text{x}]$
Answer$\lim\limits_{\text{x}\rightarrow1}\ [\text{x}]$
$\lim\limits_{\text{x}\rightarrow1^-}[\text{x}]=0$
$\lim\limits_{\text{x}\rightarrow1^-}[\text{x}]=1$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^-}\ [\text{x}]\ne\lim\limits_{\text{x}\rightarrow1^+}[\text{x}]$
Thus, $\lim\limits_{\text{x}\rightarrow1}\ [\text{x}]$ does not exist
View full question & answer→Question 222 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{logx}-\text{loga}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{logx}-\text{loga}}{\text{x}-\text{a}}$
$=\lim\limits_{\text{x}\rightarrow\text{a}}\frac{\text{log}\frac{\text{x}}{\text{a}}}{\text{a}\Big(\frac{\text{x}}{\text{a}}-1\Big)}$
Let $\text{ h}=\frac{\text{x}}{\text{a}}-1$
$=\frac{1}{\text{a}}\lim\limits_{\text{x}\rightarrow\text{a}}\frac{\text{log(h+1)}}{\text{h}}$
$=\frac{1}{\text{a}}$
View full question & answer→Question 232 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^0}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^0}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}\times\pi}{180}}{\text{x}}$ $\Big[\because\ 1^\circ=\frac{\pi}{180}\text{ radians}\Big]$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\text{x}\times\frac{\pi}{180}}\times\frac{\pi}{180}$
$=\frac{\pi}{180}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\pi\text{x}}{180}}{\frac{\text{x}\pi}{180}}$
$=\frac{\pi}{180}\times1=\frac{\pi}{180}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac{\pi}{180}$
View full question & answer→Question 242 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{5^\text{x}-1}{\sqrt{4+\text{x}}-2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5^\text{x}-1}{\sqrt{4+\text{x}}-2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(5^\text{x}-1\big){\big(\sqrt{4+\text{x}}+2}\big)}{\big(\sqrt{4+\text{x}}-2\big)\big(\sqrt{4+\text{x}}+2\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(5^\text{x}-1\big){\big(\sqrt{4+\text{x}}+2\big)}}{\text{x}}$
$=4\text{log}5$
View full question & answer→Question 252 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{bx}-\text{e}^\text{ax}}{\text{x}}$ where 0 < a < b.
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{bx}-\text{e}^\text{ax}}{\text{x}}$
$=\text{b}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ bx}-1}{\text{bx}}-\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^ \text{ ax}-1}{\text{ax}}$
$=\text{b}-\text{a}$
View full question & answer→Question 262 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{(3\text{x}-1)(4\text{x}-2)}{(\text{x}+8)(\text{x}-1)}$
Answer$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{(3\text{x}-1)(4\text{x}-2)}{(\text{x}+8)(\text{x}-1)}$ $\Big[\text{Expression is }\frac\infty\infty\Big]$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\big(12\text{x}^2-10\text{x}+2\big)}{\big(\text{x}^2+9\text{x}-8\big)}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\Bigg(\frac{12-\frac{10}{\text{x}}+\frac{2}{\text{x}^2}}{1+\frac{9}{\text{x}}-\frac{8}{\text{x}^2}}\Bigg)$
$=\frac{12-0+0}{1+0-1}$
$=12$
View full question & answer→Question 272 Marks
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}^2+7\text{x}}-\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}^2+7\text{x}}-\text{x}$
$=\lim\limits_{\text{x}\rightarrow\infty}\Bigg(\frac{\big(\sqrt{\text{x}^2+7\text{x}}-\text{x}\big)\big(\sqrt{\text{x}^2+7\text{x}}+\text{x}\big)}{\sqrt{\text{x}^2+7\text{x}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow\infty}\Bigg(\frac{\big({\text{x}^2+7\text{x}}\big)-{\text{x}}}{\sqrt{\text{x}^2+7\text{x}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{{7\text{x}}}{\sqrt{\text{x}^2+7\text{x}}+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{{7}}{\sqrt{\frac{\text{x}^2}{\text{x}^2}\frac{7\text{x}}{\text{x}^2}}+1}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{{7}}{\sqrt{1+\frac{7}{\text{x}}}+1}$
$=\frac72$
View full question & answer→Question 282 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-4\sin^3\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-4\sin^3\text{x}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{{\text{x}}}$ $\big[\because\sin3\text{x}=3\sin\text{x}-4\sin^3\text{x}\big]$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{{3\text{x}}}\times3$
$=3\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{{3\text{x}}}$
$=3\times\lim\limits_{3\text{x}\rightarrow0}\frac{\sin3\text{x}}{{3\text{x}}}$ $\big[\because\text{x}\rightarrow0,3\text{x}\rightarrow0\big]$
$=3\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=3$
View full question & answer→Question 292 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{9^\text{x}-2.6^\text{x}+4^\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{9^\text{x}-2.6^\text{x}+4^\text{x}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(3^\text{x}\big)^2-2.3^\text{x}2^\text{x}+\big(2^\text{x}\big)^2}{\text{x}^2} $
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3^\text{x}-2^\text{x}}{\text{x}}\Big)^2$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\Big(\frac{2^\text{x}-1}{\text{x}}\Big)\Big)^2$
$=\Big(\text{log}\frac{3}{2}\Big)^2$
View full question & answer→Question 302 Marks
Show that $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{|\text{x}|}$ does not exist.
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{|\text{x}|}$
We know that $|\text{x}|=\begin{cases}\text{x}, &\text{if }\text{x }\ge0\\-\text{x}, &\text{if } \text{x} < 0\end{cases}$
$\therefore\ \lim\limits_{\text{x}\rightarrow0^+}\frac{\text{x}}{|\text{x}|}=\lim\limits_{\text{x}\rightarrow0^+}\frac{\text{x}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^+}1=1$
Also, $\lim\limits_{\text{x}\rightarrow\sigma}\frac{\text{x}}{|\text{x}|}=\lim\limits_{\text{x}\rightarrow\sigma}\frac{\text{x}}{-\text{x}}=\lim\limits_{\text{x}\rightarrow\sigma}-1=-1$
$\Rightarrow\text{L.H.L}\text{ of f(x)}\ne\text{R.H.L of f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{|\text{x}|}$ does not exist.
View full question & answer→Question 312 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ 3+x}-\sin\text{x}-\text{e}^3}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ 3+x}-\sin\text{x}-\text{e}^3}{\text{x}}$
$=\text{e}^3\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$
$=\text{e}^3 \text{log e}-1$
$=\text{e}^3 -1$
View full question & answer→Question 322 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}$
$=\lim\limits_{\tan\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}$
$=1$
View full question & answer→Question 332 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{\tan3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{\tan3\text{x}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}{\sin5\text{x}}{}}{\lim\limits_{3\text{x}\rightarrow0}{\tan3\text{x}}}$
$=\frac{\lim\limits_{5\text{x}\rightarrow0}\frac{\sin5\text{x}}{5\text{x}}\times5}{\lim\limits_{3\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\times3}$ $[\because$ if x → 0 then 3x → 0 also 5x → 0$]$
$=\frac{5}{3}\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\text{ also }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{5}{3}$
View full question & answer→Question 342 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{\sin\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{\sin\text{x}^2}$
$=\frac{1}{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}}{}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac11$
$=1$
View full question & answer→Question 352 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\cos\text{x}+\sin\text{x}}{\text{x}^2+\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\cos\text{x}+\sin\text{x}}{\text{x}^2+\tan\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\cos\text{x}+\frac{\sin\text{x}}{\text{x}}}{\text{x}+\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}\cos\text{x}+\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x} \rightarrow0}\text{x}+\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{1+1}{0+1}=\frac21$
$=2$
View full question & answer→Question 362 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\Big(\frac{\pi}{2}-\text{x}\Big)\tan\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\Big(\frac{\pi}{2}-\text{x}\Big)\tan\text{x}$
Let $\text{y} =\frac{\pi}{2}-\text{x}$ as $\text{x}\rightarrow\frac{\pi}{2},\text{y}\rightarrow0$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\Big(\frac{\pi}{2}-\text{x}\Big)\tan\text{x}$
$=\lim\limits_{\text{y}\rightarrow0}\text{ y }\tan\Big(\frac{\pi}{2}-\text{y}\Big)$
$=\lim\limits_{\text{y}\rightarrow0}\text{ y }\frac{\sin\big(\frac{\pi}{2}-\text{y}\big)}{\cos\big(\frac{\pi}{2}-\text{y}\big)}$
$=\lim\limits_{\text{y}\rightarrow0}\text{ y }\frac{\cos\text{y}}{\sin\text{y}}$
$=\lim\limits_{\text{y}\rightarrow0}\cos\text{ y }\lim\limits_{\text{y}\rightarrow0}\frac{\text{y}}{\sin\text{y}}$
$=1$
View full question & answer→Question 372 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ x }-\text{e }^{\sin\text{x}}}{\text{x}-\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ x }-\text{e }^{\sin\text{x}}}{\text{x}-\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0 }\text{e}^{ \sin \text{x}}\Big[\frac{\text{e}^{\text{ x }-\sin\text{x}}-1}{\text{x}-\sin\text{x}}\Big]$
$=1\times\text{log e}$
$=1$
View full question & answer→Question 382 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\cos^2\text{x}}{1-\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\cos^2\text{x}}{1-\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{1-\sin^2\text{x}}{1-\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{(1-\sin\text{x})(1+\sin\text{x})}{(1-\sin\text{x})}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}{(1+\sin\text{x})}$
$=1+\sin\frac{\pi}{2}$
$=1+1$
$=2$
View full question & answer→Question 392 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\text{e}^\text{x}-1\big)}{1-\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\text{e}^\text{x}-1\big)}{1-\cos\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\text{e}^\text{x}-1\big)}{2\sin^2\big(\frac{\text{x}}{2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{e}^\text{x}-1\big)}{2\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{4}{\begin{pmatrix}\frac{\sin\big(\frac{\text{x}}{2}\big)}{\frac{\text{x}}{2}}\end{pmatrix}^2}$
$=\frac{1}{2}\times4$
$=2$
View full question & answer→Question 402 Marks
Show that $\lim\limits_{\text{x}\rightarrow2}\text{ e}^\frac{-1}{\text{x}}$ does not exist.
Answer$\lim\limits_{\text{x}\rightarrow0}\text{ e}^{\frac{-1}{\text{x}}}$
$\lim\limits_{\text{x}\rightarrow0^+}\text{ e}^{\frac{-1}{\text{x}}}=\lim\limits_{\text{x}\rightarrow0^+}\frac{1}{\text{e}^{\frac{1}{\text{x}}}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{e}^{\frac{1}{0+\text{h}}}}=\frac{1}{\text{e}^{\frac{1}{0}}}=\frac{1}{\text{e}^{\infty}}=\frac{1}{\infty}=0$
And,
$=\lim\limits_{\text{x}\rightarrow0^-}\text{e}^{\frac{-1}{\text{x}}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{1}{\text{e}^{\frac{1}{\text{x}}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{e}^\frac{1}{0-\text{h}}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{e}^-\frac{1}{\text{h}}}=\frac{1}{\text{e}^-\frac{1}{0}}=\frac{1}{\text{e}^{-\infty}}=\text{e}^\infty=\infty$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^+}\text{e}^{\frac{-1}{\text{x}}}\ne\lim\limits_{\text{x}\rightarrow0^-}\text{e}^{\frac{-1}{\text{x}}}$
$\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-1}{\text{x}}}$ does not exist.
View full question & answer→Question 412 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0^+}\Big\{1+\tan^2\sqrt{\text{x}}\Big\}^{\frac{1}{2}\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0^+}\Big\{1+\tan^2\sqrt{\text{x}}\Big\}^{\frac{1}{2}\text{x}}$
$=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\tan^2\sqrt{\text{x}}}{2\text{x}}\Big\}}$
$=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\tan^2\sqrt{\text{x}}}{2\text{x}}\Big\}}$
$=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\sin^2\sqrt{\text{x}}}{2\text{x}\cos^2\sqrt{\text{x}}}\Big\}}$
$=\ \text{e}^{\frac{1}{2}\lim\limits_{\text{x}\rightarrow0^+}\bigg\{\Big(\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\Big)^2\bigg\}\lim\limits_{\text{x}\rightarrow0^+}\Big\{\frac{1}{\cos^2\sqrt{\text{x}}}\Big\}}$
$=\ \text{e}^\frac{1}{2}$
$=\ \sqrt{\text{e}}$
View full question & answer→Question 422 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{8^\text{x}-2^\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{8^\text{x}-2^\text{x}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{8^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{2^\text{x}-1}{\text{x}}$
$=\text{log8}-\text{log2}$
$=\text{log4}$
View full question & answer→Question 432 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos\text{mx}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos\text{mx}}{\text{x}^2}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\frac{\text{mx}}{2}}{\text{x}^2}$
$=2\times\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{mx}}{2}}{\text{x}}\bigg)^2$
$=2\times\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{mx}}{2}}{\frac{\text{mx}}{2}}\bigg)^2\times\Big(\frac{\text{m}}{2}\Big)$
$=2\times\frac{\text{m}^2}{4}=\frac{\text{m}^2}{2}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac{\text{m}^2}{2}$
View full question & answer→Question 442 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}\cos\text{x}}{3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}\cos\text{x}}{3\text{x}}$
$=\frac13\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}\cos\text{x}}{{\text{x}}}{}$
$=\frac13\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{{\text{x}}}\times\lim\limits_{\text{x}\rightarrow0}\cos\text{x}$
$=\frac13\times1\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\cos0^\circ=1\Big]$
$=\frac13$
View full question & answer→Question 452 Marks
Find:
$\lim\limits_{\text{x}\rightarrow2}[\text{x}]$
Answer$\lim\limits_{\text{x}\rightarrow0}\ [\text{x}]$
$\lim\limits_{\text{x}\rightarrow2^-}[\text{x}]=1$
$\lim\limits_{\text{x}\rightarrow2^+}[\text{x}]=2$
Thus, $\lim\limits_{\text{x}\rightarrow2}\ [\text{x}]$ does not exist.
View full question & answer→Question 462 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}-1}-\frac{2}{\text{x}^2-1}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}-1}-\frac{2}{\text{x}^2-1}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}-1}-\frac{2}{(\text{x}-1)(\text{x}+1)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}+1-2}{(\text{x}-1)(\text{x}+1)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}-1}{(\text{x}-1)(\text{x}+1)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{(\text{x}+1)}\Big)$
$=\frac{1}{1+1}=\frac12$
View full question & answer→Question 472 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{2^{-\cos\text{x}}-1}{\pi\big(\text{x}-\frac{\pi}{2}\big)}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{2^{-\cos\text{x}}-1}{\pi\big(\text{x}-\frac{\pi}{2}\big)}$
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{2^{\sin\big(\text{x}-\frac{\pi}{2}\big)}-1}{\big(\text{x}-\frac{\pi}{2}\big)}\times\frac{1}{\text{x}}$
$=\frac{2}{\pi}\text{log}_e2$
View full question & answer→Question 482 Marks
Find:
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}}[\text{x}]$
Answer$\lim\limits_{\text{x}\rightarrow\frac52}\ [\text{x}]$
$\lim\limits_{\text{x}\rightarrow\frac{5^+}{2}}[\text{x}]=2$
$\lim\limits_{\text{x}\rightarrow\frac{5^-}{2}}[\text{x}]=2$
$\Rightarrow\lim\limits_{\text{x}\rightarrow\frac{5}{2}}\ [\text{x}]=2$
View full question & answer→Question 492 Marks
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{(a}+\text{x})-\log\text{a}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{(a}+\text{x})-\log\text{a}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{}\big(1+\frac{\text{x}}{\text{a}}\big)}{\text{a}\big(\frac{\text{x}}{\text{a}}\big)}$
$=\frac{1}{\text{a}}$
View full question & answer→Question 502 Marks
Show the $\lim\limits_{\text{x}\rightarrow0}\frac{1}{\text{x}}$ does not exist.
AnswerLet $\text{f(x)}=\frac{1}{\text{x}},$ this function is defined for every value of x excepts at x = 0
As $\text{x}\rightarrow0^+,\frac{1}{\text{x}}\rightarrow\infty$
As $\text{x}\rightarrow0^-,\frac{1}{\text{x}}\rightarrow-\infty$
$\therefore\lim\limits_{\text{x}\rightarrow0}\frac{1}{\text{x}}$ does not exist.
View full question & answer→