lim _{x rightarrow 1} (1 + log _{e} x)^{1 / l | vedclass

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Question

(B) Let $L = \lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$.As $x \rightarrow 1$,$\log _{e} x \rightarrow 0$,so the expression takes the

Vedclass · Accepted Answer

$e$

Vedclass · Answer

(B) Let $L = \lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$.As $x \rightarrow 1$,$\log _{e} x \rightarrow 0$,so the expression takes the form $1^{\infty}$.We use the standard limit formula: $\lim _{u \rightarrow 0} (1 + u)^{1/u} = e$.Let $u = \log _{e} x$. As $x \rightarrow 1$,$u \rightarrow 0$.Substituting this into the limit,we get $\lim _{u \rightarrow 0} (1 + u)^{1/u} = e$.Therefore,the correct option is $B$.

$\lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$ is equal to

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