Find this limit without using L'Hôpitale

Question

Can any give me a hint to solve this limit
$ \lim_{x\to 1} \frac{1}{x-1}-\frac{1}{ln(x)} $
I used L'Hôpitale rule and I found that the solution is -$\frac{1}{2}$ And I tried another ways like
$ T=x-1$
$\lim_{t\to 0} \frac{ln(t+1)-t}{tln(t+1)} $
And I don't get any thing after that
Any help would be very good
Note: taylor series isn't available Thanks so much :)

Answer 1

Let $t=e^x-1$ then you want to find (I changed the order here, so I get the limit $\frac{1}{2}$ instead of $-\frac{1}{2}$, sorry)

$$L=\lim\limits_{x\to 0} \frac{e^x-x-1}{x(e^x-1)}$$

multiply by $$1=\lim\limits_{x\to 0} \frac{e^x-1}{x}$$ and the problem simplifies to

$$L=\lim\limits_{x\to 0} \frac{e^x-x-1}{x^2}$$ Note that $$L=\lim\limits_{x\to 0} \frac{e^{-x}+x-1}{x^2}$$ thus

$$2L=\lim\limits_{x\to 0} \frac{e^{-x}+e^x-2}{x^2}= \lim\limits_{x\to 0} \left(\frac{1-e^{x}}{x}\right)\left(\frac{e^{-x}-1}{x}\right)=(-1)(-1)=1$$

So $$L=\frac{1}{2}$$