What is $\lim_{x→1^{-}} (1 – x^2)^{\frac{1}{\log(1 – x)}}$?

Question

The question was from a set of questions for mathematics pg entrance examination. $$\lim_{x→1^{-}} (1 – x^2)^{\dfrac{1}{\log(1 – x)}}$$

The correct answer that was given was $e$

$\log L=\lim_{x→1^{-}}\dfrac{\log(1 – x^2)}{\log(1 – x)}$

I am stuck here, since $\log 0$ is undefined, I believe I cannot use L Hopital's rule.How can I solve this question?

Answer 1

Notice that

$$\frac{\ln(1-x^2)}{\ln(1-x)}=\frac{\ln[(1-x)(1+x)]}{\ln(1-x)}=\frac{\ln(1-x)+\ln(1+x)}{\ln(1-x)}=1+\underbrace{\frac{\ln(1+x)}{\ln(1-x)}}_{\to0}\to1$$

as $x\to1^-$, and so, as this is the logarithm of the limit, it follows by continuity that the limit equals $e^1=e$.

Answer 2

As commented by Lorago, "$\ln0$ doesn't have to be defined for L'Hôpital's rule to be applicable, as long as it has a limit (in this case $−∞$)".

More precisely, by the second case of L'Hôpital's rule, as $x\to1^-,$ $$\frac{(\ln(1-x^2))'}{(\ln(1-x))'}=\frac{\frac{-2x}{1-x^2}}{\frac{-1}{1-x}}=\frac{2x}{1+x}\to1,$$ hence $\ln L=1$ and $L=e^1=e.$