Question 12 Marks
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\tan^23\text{x}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\tan^23\text{x}}{\text{x}^2}$ $=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{\text{x}}\Big)^2$ $=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\Big)^2$ $=1\times9$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=9$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\sin\text{kx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\sin\text{kx}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\text{kx}\frac{\sin\text{kx}}{\text{kx}}}$ $=\frac{1}{\text{x}}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\big(\text{a}^\text{mx}-\text{b}^\text{nx}\big)}{\text{x}}}{\frac{\sin\text{kx}}{\text{kx}}}$ $=\frac{1}{\text{x}}\text{log}\frac{\text{a}^\text{m}}{\text{b}^\text{n}}$
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Evaluate the following limits: $\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}}{\sqrt{4\text{x}^2+1}-1}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}}{\sqrt{4\text{x}^2+1}-1}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{1}{\sqrt{4+\frac{1}{\text{x}^2}}-\frac{1}{\text{x}}}$ $=\frac{1}{\sqrt{4}-0}$ $=\frac12$
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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.
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-\text{x}-1}{2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-\text{x}-1}{2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{2}-1$ $=1-1$ $=0$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{log(2+x)}+\text{log0.5}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{log(2+x)}+\text{log0.5}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}\Big(1+\frac{\text{x}}{2}\Big)}{2\Big(\frac{\text{x}}{2}\Big)}$ $=\frac{1}{2}$
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Evaluate the following one sided limits: $\lim\limits_{\text{x}\rightarrow0^-}(1+\text{cosec }\text{x})$
Answer$\lim\limits_{\text{x}\rightarrow0^-}(1+\text{cosec }\text{x})$ $=\lim\limits_{\text{x}\rightarrow0^-}1+\text{cosec }(0-\text{h})$ $=\lim\limits_{\text{h}\rightarrow0}\ 1-\text{cosec}\text{ h}$ $=\lim\limits_{\text{h}\rightarrow0}1-\frac{1}{\sin\text{h}}$ $\Rightarrow1-\frac{1}{0}=-\infty$
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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$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}+2\text{x}}{3\text{x}+2\tan3\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{3\frac{\sin\text{x}}{\text{x}}+2}{3+2\tan\frac{3\text{x}}{\text{x}}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{3\sin2\text{x}}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}2}{\lim\limits_{\text{x}\rightarrow0}3+\lim\limits_{\text{x}\rightarrow0}\frac{2\tan3\text{x}}{\text{x}}}$ $=\frac{\Big(3\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\Big)+2}{3+\Big(2\times\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\times3\Big)}$ $=\frac{(3\times2)+2}{3+(2\times3)}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ also }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{8}{9}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x+2}-\text{e}^2}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x+2}-\text{e}^2}{\text{x}}$ $=\text{e}^2\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}$ $=\text{e}^2$
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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}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{a}^\text{h}-1\big)}{\text{h}}$ $=\text{log a}$
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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$\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{4}{\text{x}^3-2\text{x}^2}\Big)$ $=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{1}{\text{x}-2}-\frac{4}{\text{x}^2(\text{x}-2)}\Big)$ $=\lim\limits_{\text{x}\rightarrow2}\Big(\frac{\text{x}^2-4}{\text{x}^2(\text{x}-2)}\Big)$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}^2(\text{x}-2)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}+2)}{\text{x}^2}$ $=\frac{(2+2)}{2^2}=\frac{4}{4}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sin2\text{x}}{\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{2}}}\frac{\sin2\text{x}}{\cos\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{2\sin\text{x}\cos\text{x}}{\cos\text{x}}$ $=2\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\sin\text{x}$ $=2\times\sin\frac{\pi}{2}$ $=2\times1$ $=2$
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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$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3\text{x}^2-4\text{x}^2+6\text{x}-1}{2\text{x}^3+\text{x}^2-5\text{x}+7}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3-\frac{4}{\text{x}}+\frac{6}{\text{x}^2}-\frac{1}{\text{x}^3}}{2+\frac{1}{\text{x}}-\frac{5}{\text{x}^2}+\frac{7}{\text{x}^3}}$ $=\frac{3-0+0-0}{2+0-0+0}$ $=\frac32$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{a}^{-\text{x}}-2}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{a}^{-\text{x}}-2}{\text{x}^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{2x}-\text{2a}^\text{x}-2}{\text{a}^2.\text{x}^2}$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\text{a}^\text{x}-1}{\text{x}}\Big)^2\times\frac{1}{\text{a}^\text{x}}$ $=(\text{log}_e\text{a})^2\times\frac{1}{\text{a}^0}$ $=(\text{log}_e\text{a})^2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{3\text{x}}-\text{e}^{2\text{x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{3\text{x}}-\text{e}^{2\text{x}}}{\text{x}}$ $=3\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{3\text{x}}-1\text{}}{3\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{2\text{x}}-1\text{}}{2\text{x}}$ $=3-2$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{3\tan^2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{3\tan^2\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{2\sin^2\text{x}}{3\tan^2\text{x}}$ $=\frac23\lim\limits_{\text{x} \rightarrow0}\frac{\sin^2\text{x}}{\Big(\frac{\sin^2\text{x}}{\cos^2\text{x}}\Big)}$ $=\frac23\lim\limits_{\text{x} \rightarrow0}\cos^2\text{x}$ $=\frac23$
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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 $\text{ x}-\frac{\pi}{2}=\text{h}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-\sin\text{h}}-1}{-\sin\text{h}}$ $=\lim\limits_{\sin\text{h}\rightarrow0}\frac{\text{e}^{-\sin\text{h}}-1}{-\sin\text{h}}$ $=1$
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Evaluate the following one sided limits: $\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}-3}{\text{x}^2-4}.$
Answer$\lim\limits_{\text{x}\rightarrow2^+}\frac{\text{x}-3}{\text{x}^2-4}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(2+\text{h})-3}{(2+\text{h)}^2-2^2}$ $\Big[\because\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h)}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(2-3+\text{h})}{(2+\text{h}-2)(2+\text{h}+2)}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{h}-1)}{(\text{h})(4+\text{h})}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1-\frac{1}{\text{h}}}{4+\text{h}}$ $\frac{1-\frac{1}{0}}{4}=-\infty$
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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$\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{1-\cos6\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{2\sin^23\text{x}}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$ $\big(1-\cos2\theta=2\sin^2\theta\big)$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sqrt{2}\sin3\text{x}}{\sqrt{2}\big(\frac{\pi}{3}-\text{x}\big)}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{3}}\frac{\sin3\text{x}}{\big(\frac{\pi}{3}-\text{x}\big)}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\sin3\big(\frac{\pi}{3}+\text{h}\big)}{\frac{\pi}{3}-\big(\frac{\pi}{3}-\text{x}\big)}$ $\big(\text{Put x}=\frac{\pi}{3}+\text{h}\big)$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(\pi+3\text{h})}{-\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-\sin3\text{h}}{-\text{h}}$ $[\sin(\pi+\theta)=-\sin\theta]$ $=3\times\lim\limits_{\text{h}\rightarrow0}\frac{\sin3\text{h}}{3\text{h}}$ $3\times1$ $\Big(\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big)$ $=3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}$ $=\text{log e }\times 1$ $=1$
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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
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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 $\text{ h}=\frac{\text{x}}{\text{a}}-1$ $=\frac{1}{\text{a}}\lim\limits_{\text{x}\rightarrow\text{a}}\frac{\text{log(h+1)}}{\text{h}}$ $=\frac{1}{\text{a}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^0}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^0}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{x}\times\pi}{180}}{\text{x}}$ $\Big[\because\ 1^\circ=\frac{\pi}{180}\text{ radians}\Big]$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\pi\text{x}}{180}}{\text{x}\times\frac{\pi}{180}}\times\frac{\pi}{180}$ $=\frac{\pi}{180}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\pi\text{x}}{180}}{\frac{\text{x}\pi}{180}}$ $=\frac{\pi}{180}\times1=\frac{\pi}{180}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\frac{\pi}{180}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{5^\text{x}-1}{\sqrt{4+\text{x}}-2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5^\text{x}-1}{\sqrt{4+\text{x}}-2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(5^\text{x}-1\big){\big(\sqrt{4+\text{x}}+2}\big)}{\big(\sqrt{4+\text{x}}-2\big)\big(\sqrt{4+\text{x}}+2\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(5^\text{x}-1\big){\big(\sqrt{4+\text{x}}+2\big)}}{\text{x}}$ $=4\text{log}5$
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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$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{bx}-\text{e}^\text{ax}}{\text{x}}$ $=\text{b}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ bx}-1}{\text{bx}}-\text{a}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^ \text{ ax}-1}{\text{ax}}$ $=\text{b}-\text{a}$
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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$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{(3\text{x}-1)(4\text{x}-2)}{(\text{x}+8)(\text{x}-1)}$ $\Big[\text{Expression is }\frac\infty\infty\Big]$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{\big(12\text{x}^2-10\text{x}+2\big)}{\big(\text{x}^2+9\text{x}-8\big)}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\Bigg(\frac{12-\frac{10}{\text{x}}+\frac{2}{\text{x}^2}}{1+\frac{9}{\text{x}}-\frac{8}{\text{x}^2}}\Bigg)$ $=\frac{12-0+0}{1+0-1}$ $=12$
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Evaluate the following limits: $\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}^2+7\text{x}}-\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}^2+7\text{x}}-\text{x}$ $=\lim\limits_{\text{x}\rightarrow\infty}\Bigg(\frac{\big(\sqrt{\text{x}^2+7\text{x}}-\text{x}\big)\big(\sqrt{\text{x}^2+7\text{x}}+\text{x}\big)}{\sqrt{\text{x}^2+7\text{x}}+\text{x}}\Bigg)$ $=\lim\limits_{\text{x}\rightarrow\infty}\Bigg(\frac{\big({\text{x}^2+7\text{x}}\big)-{\text{x}}}{\sqrt{\text{x}^2+7\text{x}}+\text{x}}\Bigg)$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{{7\text{x}}}{\sqrt{\text{x}^2+7\text{x}}+\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{{7}}{\sqrt{\frac{\text{x}^2}{\text{x}^2}\frac{7\text{x}}{\text{x}^2}}+1}$ $=\lim\limits_{\text{x}\rightarrow\infty}\frac{{7}}{\sqrt{1+\frac{7}{\text{x}}}+1}$ $=\frac72$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-4\sin^3\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-4\sin^3\text{x}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{{\text{x}}}$ $\big[\because\sin3\text{x}=3\sin\text{x}-4\sin^3\text{x}\big]$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{{3\text{x}}}\times3$ $=3\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{{3\text{x}}}$ $=3\times\lim\limits_{3\text{x}\rightarrow0}\frac{\sin3\text{x}}{{3\text{x}}}$ $\big[\because\text{x}\rightarrow0,3\text{x}\rightarrow0\big]$ $=3\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=3$
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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} $ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3^\text{x}-2^\text{x}}{\text{x}}\Big)^2$ $=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\Big(\frac{2^\text{x}-1}{\text{x}}\Big)\Big)^2$ $=\Big(\text{log}\frac{3}{2}\Big)^2$
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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.
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ 3+x}-\sin\text{x}-\text{e}^3}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ 3+x}-\sin\text{x}-\text{e}^3}{\text{x}}$ $=\text{e}^3\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $=\text{e}^3 \text{log e}-1$ $=\text{e}^3 -1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}$ $=\lim\limits_{\tan\text{x}\rightarrow0}\frac{\text{e}^{\tan\text{x}}-1}{\tan\text{x}}$ $=1$
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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$]$ $=\frac{5}{3}\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\text{ also }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{5}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{\sin\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{\sin\text{x}^2}$ $=\frac{1}{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}}{}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\frac11$ $=1$
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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$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\cos\text{x}+\sin\text{x}}{\text{x}^2+\tan\text{x}}$ $=\lim\limits_{\text{x} \rightarrow0}\frac{\cos\text{x}+\frac{\sin\text{x}}{\text{x}}}{\text{x}+\frac{\tan\text{x}}{\text{x}}}$ $=\frac{\lim\limits_{\text{x} \rightarrow0}\cos\text{x}+\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x} \rightarrow0}\text{x}+\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}}$ $=\frac{1+1}{0+1}=\frac21$ $=2$
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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 $\text{x}\rightarrow\frac{\pi}{2},\text{y}\rightarrow0$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\Big(\frac{\pi}{2}-\text{x}\Big)\tan\text{x}$ $=\lim\limits_{\text{y}\rightarrow0}\text{ y }\tan\Big(\frac{\pi}{2}-\text{y}\Big)$ $=\lim\limits_{\text{y}\rightarrow0}\text{ y }\frac{\sin\big(\frac{\pi}{2}-\text{y}\big)}{\cos\big(\frac{\pi}{2}-\text{y}\big)}$ $=\lim\limits_{\text{y}\rightarrow0}\text{ y }\frac{\cos\text{y}}{\sin\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\cos\text{ y }\lim\limits_{\text{y}\rightarrow0}\frac{\text{y}}{\sin\text{y}}$ $=1$
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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$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{ x }-\text{e }^{\sin\text{x}}}{\text{x}-\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0 }\text{e}^{ \sin \text{x}}\Big[\frac{\text{e}^{\text{ x }-\sin\text{x}}-1}{\text{x}-\sin\text{x}}\Big]$ $=1\times\text{log e}$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\cos^2\text{x}}{1-\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\cos^2\text{x}}{1-\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{1-\sin^2\text{x}}{1-\sin\text{x}}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{(1-\sin\text{x})(1+\sin\text{x})}{(1-\sin\text{x})}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}{(1+\sin\text{x})}$ $=1+\sin\frac{\pi}{2}$ $=1+1$ $=2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\text{e}^\text{x}-1\big)}{1-\cos\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\text{e}^\text{x}-1\big)}{1-\cos\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\text{e}^\text{x}-1\big)}{2\sin^2\big(\frac{\text{x}}{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{e}^\text{x}-1\big)}{2\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{4}{\begin{pmatrix}\frac{\sin\big(\frac{\text{x}}{2}\big)}{\frac{\text{x}}{2}}\end{pmatrix}^2}$ $=\frac{1}{2}\times4$ $=2$
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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.
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0^+}\Big\{1+\tan^2\sqrt{\text{x}}\Big\}^{\frac{1}{2}\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0^+}\Big\{1+\tan^2\sqrt{\text{x}}\Big\}^{\frac{1}{2}\text{x}}$ $=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\tan^2\sqrt{\text{x}}}{2\text{x}}\Big\}}$ $=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\tan^2\sqrt{\text{x}}}{2\text{x}}\Big\}}$ $=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\sin^2\sqrt{\text{x}}}{2\text{x}\cos^2\sqrt{\text{x}}}\Big\}}$ $=\ \text{e}^{\frac{1}{2}\lim\limits_{\text{x}\rightarrow0^+}\bigg\{\Big(\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\Big)^2\bigg\}\lim\limits_{\text{x}\rightarrow0^+}\Big\{\frac{1}{\cos^2\sqrt{\text{x}}}\Big\}}$ $=\ \text{e}^\frac{1}{2}$ $=\ \sqrt{\text{e}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{8^\text{x}-2^\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{8^\text{x}-2^\text{x}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{8^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{2^\text{x}-1}{\text{x}}$ $=\text{log8}-\text{log2}$ $=\text{log4}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos\text{mx}}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos\text{mx}}{\text{x}^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\frac{\text{mx}}{2}}{\text{x}^2}$ $=2\times\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{mx}}{2}}{\text{x}}\bigg)^2$ $=2\times\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\frac{\text{mx}}{2}}{\frac{\text{mx}}{2}}\bigg)^2\times\Big(\frac{\text{m}}{2}\Big)$ $=2\times\frac{\text{m}^2}{4}=\frac{\text{m}^2}{2}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\frac{\text{m}^2}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}\cos\text{x}}{3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}\cos\text{x}}{3\text{x}}$ $=\frac13\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}\cos\text{x}}{{\text{x}}}{}$ $=\frac13\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{{\text{x}}}\times\lim\limits_{\text{x}\rightarrow0}\cos\text{x}$ $=\frac13\times1\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\cos0^\circ=1\Big]$ $=\frac13$
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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.
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}-1}-\frac{2}{\text{x}^2-1}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}-1}-\frac{2}{\text{x}^2-1}\Big)$ $=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}-1}-\frac{2}{(\text{x}-1)(\text{x}+1)}\Big)$ $=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}+1-2}{(\text{x}-1)(\text{x}+1)}\Big)$ $=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}-1}{(\text{x}-1)(\text{x}+1)}\Big)$ $=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{(\text{x}+1)}\Big)$ $=\frac{1}{1+1}=\frac12$
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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$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{2^{-\cos\text{x}}-1}{\pi\big(\text{x}-\frac{\pi}{2}\big)}$ $=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{2^{\sin\big(\text{x}-\frac{\pi}{2}\big)}-1}{\big(\text{x}-\frac{\pi}{2}\big)}\times\frac{1}{\text{x}}$ $=\frac{2}{\pi}\text{log}_e2$
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Find: $\lim\limits_{\text{x}\rightarrow\frac{5}{2}}[\text{x}]$
Answer$\lim\limits_{\text{x}\rightarrow\frac52}\ [\text{x}]$ $\lim\limits_{\text{x}\rightarrow\frac{5^+}{2}}[\text{x}]=2$ $\lim\limits_{\text{x}\rightarrow\frac{5^-}{2}}[\text{x}]=2$ $\Rightarrow\lim\limits_{\text{x}\rightarrow\frac{5}{2}}\ [\text{x}]=2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{(a}+\text{x})-\log\text{a}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{(a}+\text{x})-\log\text{a}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{}\big(1+\frac{\text{x}}{\text{a}}\big)}{\text{a}\big(\frac{\text{x}}{\text{a}}\big)}$ $=\frac{1}{\text{a}}$
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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.
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}-\sin\text{x}}{\tan\text{x}+\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{2\text{x}-\sin\text{x}}{\tan\text{x}+\text{x}}$Dividing each term by x
$=\lim\limits_{\text{x}\rightarrow0}\frac{2-\frac{\sin\text{x}}{\text{x}}}{\frac{\tan\text{x}}{\text{x}}+1}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}2-\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}+1}$
$=\frac{2-1}{1+1}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{1}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{3^{2+\text{x}}-9}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{3^{2+\text{x}}-9}{\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{3^2.3^\text{x}-9}{\text{x}}$ $=9\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-1}{\text{x}}$ $=9 \text{log}_\text{e}3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{(3}+\text{x})-\log\text{(3}-\text{x})}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\log\text{(3}+\text{x})-\log\text{(3}-\text{x})}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}\Big(\frac{3+\text{x}}{3-\text{x}}\Big)}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}\Big(1+\frac{2\text{x}}{3-\text{x}}\Big)}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}\Big(1+\frac{2\text{x}}{3-\text{x}}\Big)}{\frac{2\text{x}}{3-\text{x}}}\times\lim\limits_{\text{x}\rightarrow0}\frac{2}{3-\text{x}}$ $=\frac{2}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3+1}{{\text{x}+1}}$
Answer$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^3+1}{{\text{x}+1}}$ $=\lim\limits_{\text{x}\rightarrow-1}\frac{(\text{x}+1)\big(\text{x}^2-\text{x}+1\big)}{(\text{x}+1)}$ $\Big[\text{a}^3+\text{b}^3=(\text{a}+\text{b})\big(\text{a}^2+\text{b}^2-\text{ab}\big)\Big]$ $=\lim\limits_{\text{x}\rightarrow-1}{(\text{x}^2-\text{x}-1)}$ $=(-1)^2-(-1)+1$ $=1+1+1$ $=3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{7\text{x}\cos\text{x}-3\sin\text{x}}{4\text{x}+\tan\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{7\text{x}\cos\text{x}-3\sin\text{x}}{4\text{x}+\tan\text{x}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}{7\cos\text{x}-\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}}{\text{x}}}{}}{\lim\limits_{\text{x}\rightarrow0}4+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$ $=\frac{7\times\lim\limits_{\text{x}\rightarrow0}\cos\text{x}-3\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}{4+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$ $=\frac{7\times1-3\times1}{4+1}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\text{ also }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{4}{5}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-\text{b}^\text{nx}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\text{m}\frac{\text{a}^\text{nx}-1}{\text{mx}}-\lim\limits_{\text{x}\rightarrow0}\text{n}\frac{\text{b}^\text{nx}-1}{\text{nx}}$ $=\text{mlog a}-\text{nlog b}$ $=\text{log}\Big(\frac{\text{a}^\text{m}}{\text{b}^\text{n}}\Big)$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{b}^\text{x}-2}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{b}^\text{x}-2}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{\text{b}^\text{x}-1}{\text{x}}$ $=\text{loga}+\text{logb}$ $=\text{log (ab)}$
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Evaluate the following limit: $\lim\limits_{{\text{x}}\rightarrow\frac{\pi}{2}}\frac{\text{a}^{\cot\text{x}}-\text{a}^{\cos\text{x}}}{\cot\text{x}-\cos\text{x}}$
Answer$\lim\limits_{{\text{x}}\rightarrow\frac{\pi}{2}}\frac{\text{a}^{\cot\text{x}}-\text{a}^{\cos\text{x}}}{\cot\text{x}-\cos\text{x}}$ $=\lim\limits_{{\text{x}}\rightarrow\frac{\pi}{2}}\text{a}^{\cos\text{x}}\Big[\frac{\text{a}^{\cot\text{x}-{\cos\text{x}}}-1}{\cot\text{x}-\cos\text{x}}\Big]$ $=1\times\text{log a}$ $=\text{log a}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{b}^\text{x}+\text{c}^\text{x}-3}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{b}^\text{x}+\text{c}^\text{x}-3}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{\text{b}^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{\text{c}^\text{x}-1}{\text{x}}$ $=\text{log a}+\text{log b}+\text{log c}$ $=\text{log (abc)}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{b}^\text{x}-\text{c}^\text{x}-\text{d}^\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}+\text{b}^\text{x}-\text{c}^\text{x}-\text{d}^\text{x}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\text{c}^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\text{d}^\text{x}-1}{\text{x}}$ $=\text{log a}+\text{log b}-\text{log c}-\text{log d}$ $=\text{log}\Big(\frac{\text{ab}}{\text{cd}}\Big)$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\log\big|1+\text{x}^3\big|}{\sin^3\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\log\big|1+\text{x}^3\big|}{\sin^3\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\log\big|1+\text{x}^3\big|}{\sin^3\text{x}}\times\frac{1}{\lim\limits_{\text{x}\rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^3}$ $=1\times1$ $=1$
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Evaluate the following one sided limits: $\lim\limits_{\text{x}\rightarrow\frac{-\pi}{2^+}}\sec\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi^+}{2}}\sec\text{x}$ $=\lim\limits_{\text{h}\rightarrow0}\sec\big(-\frac{\pi}{2}+\text{h}\big)$ $=\sec\Big(-\frac{\pi}{2}+0\Big)$ $=\sec\Big(-\frac{\pi}{2}\Big)$ $=\frac{1}{\cos\Big(-\frac\pi2\Big)}$ $=\frac{-1}{\Big(\cos\frac{\pi}{2}\Big)}$ $\Rightarrow\frac{-1}{0}=-\infty$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{sinx}-1}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{sinx}-1}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{sinx}-1}{\text{sinx}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{sinx}}{\text{x}}$ $=\text{log e}\times1$ $=1$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-1}{\text{b}^\text{nx}-1},\text{n}\not=0$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-1}{\text{b}^\text{nx}-1},\text{n}\not=0$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{mx}-1}{\text{mx}}\times\frac{1}{\lim\limits_{\text{x}\rightarrow0}\frac{\text{b}^\text{nx}-1}{\text{nx}}}\times\frac{\text{m}}{\text{n}}$ $=\frac{\text{mloga}}{\text{nlogb}},\text{n}\not=0$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}(1+\text{x})}{3^{\text{x}}-1 }$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}(1+\text{x})}{3^{\text{x}}-1 }$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}(1+\text{x})}{\text{x}}\times\frac{1}{\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-1}{\text{x}}}$ $=\frac{1}{\text{log3}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(2^\text{x}-1\big)}{1-\cos\text{x}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{e}^\text{x}}{\sin2\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{e}^\text{x}}{\sin2\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{1}}{\sin2\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-\text{1}}{\sin2\text{x}}$ $=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{2x}-\text{1}}{2\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{2x}}{\sin2\text{x}}\Big)-\frac{1}{2}\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{2x}}{\sin2\text{x}}\Big)$ $=1-\frac{1}{2}$ $=\frac12$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\text{x}}-\text{e}^5}{\text{x}-5}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\text{x}}-\text{e}^5}{\text{x}-5}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{5+\text{h}}-\text{e}^5}{\text{h}}$ $=\text{e}^5\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^{\text{h}}-1}{\text{h}}$ $=\text{e}^5\times1$ $=\text{e}^5$
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Evaluate the following limit: $\lim\limits_{\theta\rightarrow0}\frac{1-\cos4\theta}{1-\cos6\theta}$
Answer$\lim\limits_{\theta\rightarrow0}\frac{1-\cos4\theta}{1-\cos6\theta}$ $=\lim\limits_{\theta \rightarrow0}\frac{2\sin^22\theta}{2\sin^23\theta}$ $=\lim\limits_{\theta \rightarrow0}\frac{(\sin2\theta)^2}{(\sin3\theta)^2}$ $=\frac{\lim\limits_{\theta \rightarrow0}\big(\frac{\sin2\theta}{2\theta}\big)^2\times4\theta^2}{\lim\limits_{\theta \rightarrow0}\big(\frac{\sin3\theta}{3\theta}\big)^2\times9\theta^2}$ $=\frac{1\times4\theta^2}{1\times9\theta^2}$ $=\frac{4}{9}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{8^\text{x}-4^\text{x}-2^\text{x}+1}{\text{x}^2}$
Answer$\lim\limits_{\text{x}\rightarrow0}\bigg[\frac{8^\text{x}-4^\text{x}-2^\text{x}+1}{\text{x}^2}\bigg]$ $=\lim\limits_{\text{x}\rightarrow0}\Bigg[\frac{\big(2^\text{x}\big)^3-\big(2^\text{x}\big)^2-2^\text{x}+1}{\text{x}^2}\Bigg]$ $=\lim\limits_{\text{x}\rightarrow0}\Bigg[\frac{\big(2^\text{x}\big)^2\big(2^\text{x}-1\big)-1\big(2^\text{x}-1\big)}{\text{x}^2}\Bigg]$ $=\lim\limits_{\text{x}\rightarrow0}\Bigg[\frac{\big(2^\text{2x}-1\big)\big(2^\text{x}-1\big)}{\text{x}^2}\Bigg]$ $=\lim\limits_{\text{x}\rightarrow0}\Bigg[\frac{2\big(2^\text{2x}-1\big)}{2\text{x}}\times\Big(\frac{2^\text{x}-1}{\text{x}}\Big)\Bigg]$ $=2\text{ log }2\times\text{log }2$ $=\text{ log }(2)^2\times\text{log }2$ $=(\text{ log }4)\times(\text{log }2)$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-7\text{x}+12}{{\text{x}^2}-3\text{x}-4}$
Answer$\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-7\text{x}+12}{{\text{x}^2}-3\text{x}-4}$ $=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-3\text{x}-4\text{x}+12}{\text{x}^2+\text{x}-4\text{x}-4}$ $=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}(\text{x}-3)-4(\text{x}-3)}{\text{x}(\text{x}+1)-1(\text{x}+1)}$ $=\lim\limits_{\text{x}\rightarrow4}\frac{(\text{x}-3)(\text{x}-4)}{(\text{x}-4)(\text{x}+1)}$ $=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}-3}{\text{x}+1}$ $=\frac{4-3}{4+1}$ $=\frac15$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{log(a+x)}-\text{log(a}-\text{x})}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{log(a+x)}-\text{log(a}-\text{x})}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}\Big(\frac{\text{a}+\text{x}}{\text{a}-\text{x}}\Big)}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{log}\Big(1+\frac{2\text{x}}{\text{a}-\text{x}}\Big)}{\frac{\text{2x}}{\text{a}-\text{x}}}\times\lim\limits_{\text{x}\rightarrow0}\frac{2}{\text{a}-\text{x}}$ $=\frac{2}{\text{a}}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1+\sin\text{x}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1+\sin\text{x}}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $=\text{log e+1}$ $=2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow-\frac12}\frac{8\text{x}^3+1}{2{\text{x}}+1}$
Answer$\lim\limits_{\text{x}\rightarrow-\frac12}\frac{8\text{x}^3+1}{2{\text{x}}+1}=\lim\limits_{\text{x}\rightarrow-\frac12}\frac{8\Big(\text{x}^2+\frac18\Big)}{2\Big(\text{x}+\frac12\Big)}$ $=\frac82\lim\limits_{\text{x}\rightarrow-\frac12}\frac{\Big(\text{x}^3+\big(\frac12\big)^3\Big)}{\text{x}+\frac12}$ $=4\lim\limits_{\text{x}\rightarrow-\frac{1}{2}}\frac{\Big(\text{x}+\frac12\Big)\Big(\text{x}^2+\frac14-\frac12\text{x}\Big)}{\Big(\text{x}+\frac12\Big)}$ $=4\Big(\Big(\frac{-1}{2}\Big)+\frac14-\frac12\Big(\frac{-1}{2}\Big)\Big)$ $=4\Big(\frac14+\frac14+\frac14\Big)$ $=3$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^4-16}{{\text{x}}-2}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^4-16}{{\text{x}}-2}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{\big(\text{x}^2+4\big)\big(\text{x}^2+4\big)}{({\text{x}}-2)}$ $=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+2)\big(\text{x}^2+4\big)}{({\text{x}}-2)}$ $=\lim\limits_{\text{x}\rightarrow2}{(\text{x}+2)\big(\text{x}^2+4\big)}$ $=(2+2)(4+4)$ $=4(8)$ $=32$
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Find $\lim\limits_{\text{x}\rightarrow1}\text{f(x)}$ if $\text{f(x)}=\begin{cases}\text{x}^2-1, & \text{x} \le1\\-\text{x}^2-1, &\text{x} > 1\end{cases}.$
Answer$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^-}\text{x}^2-1=\lim\limits_{\text{h}\rightarrow0}(-1-\text{h})^2-1=1-1=0$ $\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^+}\text{x}^2-\text{x}^2-1=\lim\limits_{\text{h}\rightarrow0}-(1-\text{h})^2-1=-2$ Since, $\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}\ne\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}$ $\therefore\ \lim\limits_{\text{x}\rightarrow1}\text{f(x)}$ doesnot exist.
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Let $\text{f(x)}=\begin{cases}\text{x}+1, & \text{if x}> 0\\\text{x}-1, &\text{if x} < 0\end{cases}.$ Prove that $\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.
Answer$\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{ x}+1$ $=\lim\limits_{\text{h}\rightarrow0}(0+\text{h})+1=1$ Also, $\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}\neq\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}$ $=\lim\limits_{\text{h}\rightarrow0}(0-\text{h})-1=-1$ $\Rightarrow\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}$ Hence, limit does not exist.
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^\circ}{\text{x}^\circ}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}^\circ}{\text{x}^\circ}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{{\sin\frac{\text{x}\times\pi}{180}}}{{\text{x}}\times\frac{\pi}{180}}$ $\Big[\because1^\circ=\frac{\pi}{180}\text{ radious}\Big]$ $=\lim\limits_{{\frac{\pi\text{x}}{180}}\rightarrow0}\frac{\sin\frac{\text{x}\times\pi}{180}}{\frac{\pi\text{x}}{180}}$ $\Big[\because\text{if x}\rightarrow0\text{ then }\frac{\pi\text{x}}{180}\rightarrow0\Big]$ $=1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
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Evaluate the following one sided limits: $\lim\limits_{\text{x}\rightarrow0^+}\frac{1}{3\text{x}}.$
Answer$\lim\limits_{\text{x}\rightarrow0^+}\frac{1}{3\text{x}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{3(0+\text{h)}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{(0+3\text{h})}$ $=\frac{1}{0}=\infty$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{log(1+x)}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}}-1}{\text{log(1+x)}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}}-1\big)\big(\sqrt{1+\text{x}}+1\big)}{\log\text{(1+x)}{\big(\sqrt{1+\text{x}}+1\big)}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{log(1+x)}{\big(\sqrt{1+\text{x}}+1\big)}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{1}{\frac{\text{log(1+x)}}{\text{x}}}\times\lim\limits_{\text{x}\rightarrow0}\frac{1}{\big(\sqrt{1+\text{x}+1}\big)}$ $=1\times\frac{1}{2}$ $=\frac{1}{2}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{3}{\text{x}^2-3\text{x}}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{3}{\text{x}^2-3\text{x}}\Big)$$=\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}-3}-\frac{3}{\text{x}(\text{x}-3)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\Big(\frac{\text{x}-3}{\text{x}(\text{x}-3)}\Big)$
$=\lim\limits_{\text{x}\rightarrow3}\Big(\frac{1}{\text{x}}\Big)$
$=\frac{1}{3}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{e}^\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{e}^\text{x}-1}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{\text{2x}}\times2\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{e}^\text{x}-1}$ $=1\times2\times\text{log}\text{ e}$ $=2$
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Evaluate the following one sided limits: $\lim\limits_{\text{x}\rightarrow-8^+}\frac{2\text{x}}{\text{x}+8}.$
Answer$\lim\limits_{\text{x}\rightarrow-8^+}\frac{2\text{x}}{\text{x}+8}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{2(-8+\text{h})}{(-8+\text{h)+8}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-16+2\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-16}{\text{h}}+2$ $\Rightarrow\frac{-16}{0}+2=-\infty$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}-\text{a}^{\text{-x}}}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{x}-\text{a}^{\text{-x}}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^\text{2x}-1}{\text{xa}^\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{2\big(\text{a}^{2\text{x}}-1\big)}{2\text{x}}\lim\limits_{\text{x}\rightarrow0}\frac{1}{\text{a}^\text{x}}$ $=2\text{log}_e 2$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\text{log}_\text{a}(\text{x}-1)}$
Answer$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}-2}{\text{log}_\text{a}(\text{x}-1)}$ Let x - 2 = h $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{log}_\text{a}(\text{h+1})}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{log a}}{\text{log(h+1)}}}{\text{h}}$ $=\text{log a}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}^2+\text{x}-2}-\frac{\text{x}}{\text{x}^3-1}\Big)$
Answer$\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}^2+\text{x}-2}-\frac{\text{x}}{\text{x}^3-1}\Big)$$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}^2+\text{x}-2}-\frac{\text{x}}{(\text{x}-1)(\text{x}^2+\text{x}+1)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}^3-1-\text{x}^3-\text{x}^2+2\text{x}}{(\text{x}^3-1)(\text{x}^2+\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Bigg(\frac{\big(\text{x}^2-2\text{x}+1\big)}{\big(\text{x}^3-1\big)\big(\text{x}^2+\text{x}-2\big)}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{(\text{x}-1)(\text{x}-1)}{(\text{x}-1)(\text{x}^2+1+\text{x})(\text{x}^2+\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}-1}{(\text{x}^2+1+\text{x})(\text{x}+2)(\text{x}-1)}\Big)$
$=\frac{1}{(1+1+1)(1+2)}$
$=\frac19$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{5^\text{x}+3^\text{x}+2^\text{x}-3}{\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{5^\text{x}+3^\text{x}+2^\text{x}-3}{\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{5^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{2^\text{x}-1}{\text{x}}$ $=\text{log5}+\text{log3}+\text{log2}$ $=\text{log 30}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{5\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{5\text{x}}$ $=\frac{1}{5}\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3$ $=\frac35\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}$ $=\frac{3}{5}\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\frac{3}{5}$
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Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\text{x}-\cos\text{a}}{\text{x}-\text{a}}$
Answer$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\cos\text{x}-\cos\text{a}}{\text{x}-\text{a}}$ $=\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\big(-2\sin\big(\frac{\text{x}+\text{a}}{2}\big)\sin\big(\frac{\text{x}-\text{a}}{2}\big)\big)}{\text{x}-\text{a}}$ $=-2\lim\limits_{\text{x}\rightarrow{\text{a}}}\sin\Big(\frac{\text{x}+\text{a}}{2}\Big)\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\big(\frac{\text{x}-\text{a}}{2}\big)}{\text{x}-\text{a}}$ $=-2\times\sin\Big(\frac{\text{a}+\text{a}}{2}\Big)\times\Bigg(\lim\limits_{\text{x}\rightarrow{\text{a}\rightarrow0}}\frac{\sin\big(\frac{\text{x}-\text{a}}{2}\big)}{\frac{\text{x}-\text{a}}{2}}\Bigg)\times\frac{1}{2}$ $=-2\sin\text{a}\times1\times\frac12$ $\Big[\because\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=-\sin\text{a}$
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