mathop {lim }limits_{x to 0} frac{{{e^{sin x} | vedclass

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(A) हम जानते हैं कि $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$ और $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$.$\mathop {\lim

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$1$

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(A) हम जानते हैं कि $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$ और $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$.$\mathop {\lim }\limits_{x 	o 0} \frac{e^{\sin x} - 1}{x} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{e^{\sin x} - 1}{\sin x} 	imes \frac{\sin x}{x} ight)$$= \left( \mathop {\lim }\limits_{\sin x 	o 0} \frac{e^{\sin x} - 1}{\sin x} ight) 	imes \left( \mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} ight)$$= 1 	imes 1 = 1$.वैकल्पिक रूप से,$L$-Hospital नियम का उपयोग करते हुए:$\mathop {\lim }\limits_{x 	o 0} \frac{e^{\sin x} - 1}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(e^{\sin x} - 1)}{\frac{d}{dx}(x)}$$= \mathop {\lim }\limits_{x 	o 0} \frac{\cos x \cdot e^{\sin x}}{1} = \cos(0) \cdot e^{\sin(0)} = 1 \cdot e^0 = 1$.

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

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