Question

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

Answer 1

Let $f\left(x\right)=(1-x^2{)}^{\frac{1}{\log (1-x)}}$
For $f\left(x\right)$ to be defined, $x\in (-\infty ,1)-\left\{0\right\}$
So, $\underset{x\to 1}{lim}(1-x^2{)}^{\frac{1}{\log (1-x)}}=\underset{x\to 1^{-}}{lim}(1-x^2{)}^{\frac{1}{\log (1-x)}}$
$=e^{\underset{x\to 1^{-}}{lim}\frac{\log (1-x^2)}{\log (1-x)}}$
$=e^{\underset{x\to 1^{-}}{lim}\frac{\left(\frac{-2x}{1-x^2}\right)}{\left(\frac{-1}{1-x}\right)}}$ (Using L- Hospital rule)
$=e^{\underset{x\to 1^{-}}{lim}\frac{2x}{1+x}}$
$=e$

Alternate Solution:
$\underset{x\to 1}{lim}(1-x^2{)}^{\frac{1}{\log (1-x)}}=\underset{x\to 1}{lim}(1-x{)}^{{\log }_{(1-x)}e}\cdot (1+x{)}^{\frac{1}{\log (1-x)}}=e\cdot 2^0=e[\because y^{{\log }_{y}x}=x]$