A) 0
B) 1
C) 2
D) Non existent
Correct Answer: C
Solution :
\[y=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{-x}}}{\sin x}\] Þ \[y=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left[ 1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+.... \right]-\left[ 1-\frac{x}{1!}+\frac{{{x}^{2}}}{2!}-.... \right]}{\sin x}\] Þ \[y=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\,\left[ \frac{x}{1!}+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}+............. \right]}{\sin x}\] Þ \[y=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\,\left[ 1+\frac{{{x}^{2}}}{3!}+\frac{{{x}^{4}}}{4!}+........... \right]}{\frac{\sin x}{x}}\] Þ \[y=\frac{\underset{x\to 0}{\mathop{\lim }}\,2\,\left[ 1+\frac{{{x}^{2}}}{2!}+....... \right]}{\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}}\] Þ \[y=\frac{2}{1}=2\] Trick : Applying L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{-x}}}{\sin x}\]\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}+{{e}^{-x}}}{\cos x}=\frac{{{e}^{0}}+\frac{1}{{{e}^{0}}}}{\cos 0}=\frac{1+1}{1}=2\].
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