I have trouble finding the limit of the following : $$ \lim\limits_{x\to 1} \frac{(\ln x)^2}{1-x} $$ using the rule from L´Hopital.
Since both quotients converge to $0$, I should be able to use L´Hopitals´s rule right?
But when I do that the derivations converge to $$\frac{0}{-1}$$ Does that mean that there's no limit?
Using l'Hopital's rule, we find $$ \lim_{x \to 1} \frac {\ln^2 x}{1-x} = \lim_{x \to 1} \frac {2 \ln x \cdot \frac {1}{x}}{-1} = \lim_{x \to 1} - \frac {2 \ln x}{x} = \lim_{x \to 1} 0 = \boxed {0}. $$We know l'Hopital's rule applies because it is of the form $ \frac {\ln^2 1}{1-1} = \frac {0}{0} $, which is indeed indeterminate.
We can use approximations : $\ln(x) \approx x-1$ when $x$ is close to $1$. Hence the required limit reduces to $\lim_{x \to 1}(1-x)=0$
Here are the steps $$ \lim\limits_{x\to 1} \frac{(\ln x)^2}{1-x} = \lim\limits_{x\to 1} \frac{\frac{d}{dx}(\ln x)^2}{\frac{d}{dx}[1-x]} $$ $$ =\lim\limits_{x\to 1} \frac{2(\ln x)\frac{d}{dx}[\ln x]}{\frac{d}{dx}[1]-\frac{d}{dx}[x]} =2\lim\limits_{x\to 1} \frac{\frac{\ln x}{x}}{0-1} $$ $$ =-2\lim\limits_{x\to 1} \frac{\ln x}{x} =-2\left(\frac{0}{1}\right)=-2(0)=0$$
by L'Hospital we get $$\lim_{x \to 1}2\ln(x)\frac{1}{x}(-1)=0$$
