MCQ
$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]$ is equal to
- A$-1$
- B$0$
- ✓$1$
- DNone of these
Using $ L-$ Hospital’s rule three times, then
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}.\cos x}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}{{\cos }^2}x + \sin x.{e^{\sin x}}}}{{\sin x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}.{{\cos }^3}x + {e^{\sin x}}2\cos x\sin x + {e^{\sin x}}.\cos x\sin x + {e^{\sin x}}.\cos x}}{{\cos x}}$
$ = 1$.
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