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

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Question

(A) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{	an x}} - {e^x}}}{{	an x - x}}$.Rewrite the numerator by factoring out $e^x

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$1$

Vedclass · Answer

(A) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{	an x}} - {e^x}}}{{	an x - x}}$.Rewrite the numerator by factoring out $e^x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x}({e^{	an x - x}} - 1)}}{{	an x - x}}$.Using the property of limits $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$,where $u = 	an x - x$:As $x 	o 0$,$u = 	an x - x 	o 0$.Thus,the expression becomes:$\mathop {\lim }\limits_{x 	o 0} {e^x} 	imes \mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = {e^0} 	imes 1 = 1 	imes 1 = 1$.

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

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