Hello, I have spent several hours stuck on this limit:

$\displaystyle lim(x->0) \frac{ln(1+x)-x}{x^2}$

I managed to solve it by L'hopital, its limit is -0.5 but I was wondering if anyone can come up with a different method, as I am not supposed to use L'hopital at this stage.

I tried

$\displaystyle lim\hspace{0.1cm} ln(1+x)^{\frac{1}{x^2} } - lim \frac{1}{x} $ =$\displaystyle ln \hspace{0.1cm} lim(1+x)^{\frac{1 1}{x x}} - lim \frac{1}{x}$ =$\displaystyle ln e^{1/x} - lim \frac{1}{x}$

But it did not work.

Any idea? Thanks!