The correct option is B 1e Substitute x=x+e lim_x→ elog (e+x) log_eex =lim_x→ elog(1+x/e)xe e #160; #160; #160;(use formula) ...

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Chain Rule of Differentiation

limx→ elog x-...

Question

$lim x \to e \frac{log x - 1}{x - e} =$

1

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\frac{1}{e}

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e^{2}

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e

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Solution

The correct option is B $\frac{1}{e}$
Substitute $x = x + e$
${lim}_{x \to e} \frac{log (e + x) - {log}_{e} e}{x}$
$= {lim}_{x \to e} \frac{log (1 + \frac{x}{e})}{\frac{x}{e} e}$ (use formula)
$= \frac{1}{e}$

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