A) 0
B) \[\infty \]
C) 2
D) \[{{e}^{2}}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x({{e}^{x}}-1)}{1-\cos x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{x({{e}^{x}}-1)}{2{{\sin }^{2}}\frac{x}{2}}\] \[=2\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( \frac{x}{2} \right)}^{2}}}{{{\sin }^{2}}\frac{x}{2}}\left( \frac{{{e}^{x}}-1}{x} \right)\] \[=2\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{\frac{x}{2}}{\sin \frac{x}{2}} \right)}^{2}}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{x}}-1}{x} \right)\] \[=2.1.1=2\] Alternate Solution: \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x({{e}^{x}}-1)}{1-\cos x}\] using L Hospitals rule \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{({{e}^{x}}-1)+x{{e}^{x}}}{\sin x}\] Again using L Hospitals rule \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}+x{{e}^{x}}+{{e}^{x}}}{\cos x}=\frac{1+0+1}{1}=1\]
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