A) 0
B) \[\infty \]
C) ?2
D) 2
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{x\,({{e}^{x}}-1)}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{2x\,({{e}^{x}}-1)}{4.{{\sin }^{2}}\frac{x}{2}}\] \[=2\underset{x\to 0}{\mathop{\lim }}\,\,\left[ \frac{{{(x/2)}^{2}}}{{{\sin }^{2}}\frac{x}{2}} \right]\,\left( \frac{{{e}^{x}}-1}{x} \right)=2.\]
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