_ x 0 e^ x - 1 x = - Vidyadip

Question

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} - 1}}{x} = $

✓

Answer

a
(a) $\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\sin x}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\sin x}} - 1}}{{\sin x}} \times \frac{{\sin x}}{x}$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\sin x}} - 1}}{{\sin x}} \times \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x}}{x} = 1 \times 1 = 1$.

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\sin x}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{\cos x\,{e^{\sin x}}}}{1} = 1.\,{e^0} = 1.$

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