operatorname{Lim}_{x rightarrow 0} frac{e-(1+ | vedclass

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(A) Let $L = \operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$.Using the identity $a^b = e^{b \ln a}$,we have $(1+2x)^{\frac{1

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$e$

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(A) Let $L = \operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$.Using the identity $a^b = e^{b \ln a}$,we have $(1+2x)^{\frac{1}{2x}} = e^{\frac{\ln(1+2x)}{2x}}$.$L = \operatorname{Lim}_{x \rightarrow 0} \frac{e - e^{\frac{\ln(1+2x)}{2x}}}{x} = e \operatorname{Lim}_{x \rightarrow 0} \frac{1 - e^{\frac{\ln(1+2x)}{2x} - 1}}{x}$.Using the expansion $\ln(1+t) = t - \frac{t^2}{2} + \dots$,for $t=2x$,$\ln(1+2x) = 2x - \frac{(2x)^2}{2} + \dots = 2x - 2x^2 + \dots$.Then $\frac{\ln(1+2x)}{2x} = 1 - x + \dots$.So,$L = e \operatorname{Lim}_{x \rightarrow 0} \frac{1 - e^{-x}}{x} = e \operatorname{Lim}_{x \rightarrow 0} \frac{1 - (1 - x)}{x} = e \operatorname{Lim}_{x \rightarrow 0} \frac{x}{x} = e$.

$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :

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