TITLE: How do I take the limit with invoking L'Hospital's rule? $\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{\ln(x)}\right)$
QUESTION [5 upvotes]: Need to take the limit:
$$\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{\ln(x)}\right) = \lim_{x \to 1}\left(\frac{x\cdot \ln(x)-x+1}{(x-1)\cdot \ln(x)}\right)=(0/0)$$
Now I can use L'Hospital's rule:
$$\lim_{x \to 1}\left(\frac{1\cdot \ln(x)+x\cdot \frac{1}{x}-1}{1\cdot \ln(x)+(x-1)\cdot\frac{1}{x}}\right)= \lim_{x \to 1}\left(\frac{\ln(x)+1-1}{\ln(x)+\frac{(x-1)}{x}}\right)=\lim_{x \to 1}\left(\frac{\ln(x)}{\frac{(x-1)+x \cdot \ln(x)}{x}}\right)=\lim_{x \to 1}\frac{x\cdot \ln(x)}{x-1+x\cdot \ln(x)}=\frac{1\cdot 0}{1-1+0}=(0/0)$$
As you can see I came to $(0/0)$ again. So what I have to do to solve this problem?

REPLY [3 votes]: You're not guaranteed a solution with one application of L'Hopital's rule. Sometimes, you'll have to apply it multiple times to get the limit.