lim _{x rightarrow 0} frac{e^{tan x}-e^x}{tan | vedclass

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Question

(A) We are given the limit $L = \lim _{x ightarrow 0} \frac{e^{	an x}-e^x}{	an x-x}$.Since the form is $\frac{0}{0}$,we can use the Taylor series

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$1$

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(A) We are given the limit $L = \lim _{x ightarrow 0} \frac{e^{	an x}-e^x}{	an x-x}$.Since the form is $\frac{0}{0}$,we can use the Taylor series expansion $e^u = 1 + u + \frac{u^2}{2} + \dots$.$e^{	an x} = 1 + 	an x + \frac{	an^2 x}{2} + \dots$$e^x = 1 + x + \frac{x^2}{2} + \dots$Substituting these into the numerator:$e^{	an x} - e^x = (	an x - x) + \frac{1}{2}(	an^2 x - x^2) + \dots$Now,the limit becomes $\lim _{x ightarrow 0} \frac{(	an x - x) + \frac{1}{2}(	an^2 x - x^2)}{	an x - x} = \lim _{x ightarrow 0} [1 + \frac{1}{2} \frac{	an^2 x - x^2}{	an x - x}]$.Using $	an x \approx x + \frac{x^3}{3}$,we have $	an^2 x - x^2 \approx (x + \frac{x^3}{3})^2 - x^2 \approx x^2 + \frac{2x^4}{3} - x^2 = \frac{2x^4}{3}$.Also,$	an x - x \approx \frac{x^3}{3}$.Thus,$\frac{	an^2 x - x^2}{	an x - x} \approx \frac{2x^4/3}{x^3/3} = 2x$.As $x ightarrow 0$,this term approaches $0$.Therefore,the limit is $1 + 0 = 1$.

$\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x} = $

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