MCQ
$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]$ is equal to
  • A
    $-1$
  • B
    $0$
  • $1$
  • D
    None of these

Answer

Correct option: C.
$1$
c
(c) $\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]$,      $\left( {\frac{0}{0} \, \, \,{\rm{form}}} \right)$

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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