The correct option is A e k=lim _ x→ 1 (1 x^ 2 )^ 1 /log #160;(1 x) #160; Applying log on both the sides, log k =lim _ x→ 1 1 /log #160 ...

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Continuity of a Function

limx→ 11-x21l...

Question

$lim x \to 1 (1 - x^{2})^{\frac{1}{log (1 - x)}} =$

e

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\frac{1}{e}

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\frac{1}{e^{2}}

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e^{2}

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Solution

The correct option is A $e$
$k = {lim}_{x \to 1} (1 - x^{2})^{\frac{1}{log (1 - x)}}$
Applying log on both the sides,
$log k = {lim}_{x \to 1} \frac{1}{log (1 - x)} log (1 - x^{2})$
$k = e^{{lim}_{x \to 1} \frac{log (1 - x^{2})}{log (1 - x)}}$
It is of the form $\frac{0}{0}$ , so applying L-Hospital's rule
$k = e^{{lim}_{x \to 1} \frac{- 2 x}{(1 - x^{2}) \frac{- 1}{1 - x}}}$
$\Rightarrow k = e$

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limx→1(1−x2)1log(1−x)=

The correct option is A ek=limx→1(1−x2)1log(1−x)Applying log on both the sides,logk=limx→11log(1−x)log(1−x2)k=elimx→1log(1−x2)log(1−x)It is of the form 00, so applying L-Hospital's rulek=elimx→1−2x(1−x2)−11−x⇒k=e

$lim x \to 1 (1 - x^{2})^{\frac{1}{log (1 - x)}} =$