What is the limit of $\lim_{x\to1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$? $$\lim_{x\to1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$$
I assume that I have to calculate with the L'Hospital rule but I don't know how does it work in this case.
 A: HINT: consider $$\frac{x-1-\log(x)}{(x-1)\log(x)}$$ and use L'Hospital
A: Use L'Hospital's rule.
First, simplify the expression by writing it as $$\lim_{x\to 1} \frac{f(x)}{g(x)}$$ (where you are tasked with finding $f$ and $g$).
Then, you can calculate the limit as $$\lim_{x\to 1}\frac{f'(x)}{g'(x)}.$$
A: HINT:
$$\lim_{x\to 1}\left(\frac{1}{\ln(x)}-\frac{1}{x-1}\right)=$$
$$\lim_{x\to 1}\left(\frac{-1+x-\ln(x)}{\ln(x)(x-1)}\right)=$$
$$\lim_{x\to 1}\left(\frac{\frac{\text{d}}{\text{d}x}\left(-1+x-\ln(x)\right)}{\frac{\text{d}}{\text{d}x}\left(\ln(x)(x-1)\right)}\right)=$$
$$\lim_{x\to 1}\left(\frac{1-\frac{1}{x}}{\frac{x-1}{x}+\ln(x)}\right)=$$
$$\lim_{x\to 1}\left(\frac{x-1}{-1+x+x\ln(x)}\right)=$$
$$\lim_{x\to 1}\left(\frac{\frac{\text{d}}{\text{d}x}\left(x-1\right)}{\frac{\text{d}}{\text{d}x}\left(-1+x+x\ln(x)\right)}\right)=$$
$$\lim_{x\to 1}\left(\frac{1}{2+\ln(x)}\right)$$
You can complete it I think!
A: It is better to simplify a bit before applying L'Hospital's Rule. Thus we have
\begin{align}
L &= \lim_{x \to 1}\frac{1}{\log x} - \frac{1}{x - 1}\notag\\
&= \lim_{h \to 0}\frac{h - \log(1 + h)}{h\log(1 + h)}\text{ (putting }x = 1 + h)\notag\\
&= \lim_{h \to 0}\dfrac{h - \log(1 + h)}{h^{2}\cdot\dfrac{\log(1 + h)}{h}}\notag\\
&= \lim_{h \to 0}\frac{h - \log(1 + h)}{h^{2}}\notag\\
&= \lim_{h \to 0}\dfrac{1 - \dfrac{1}{1 + h}}{2h}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{2}\lim_{h \to 0}\frac{1}{1 + h}\notag\\
&= \frac{1}{2}\notag
\end{align}
