<s>[INST] Limlit of $\frac{x-x\ln(1+x)-\ln(1+x)}{x^3+x^2}$ at $0$ without l'Hôpital or Taylor series I have $f : x \mapsto \frac{\ln(1+x)}{x}$ which derivative is $$\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2}$$
I want to find the limit as $x$ goes to $0$ of this derivative. I've tried simplifying the expression to $$\frac{x - x\ln(1+x) - \ln(1+x)}{x^3 + x^2}$$ but I'm still struggling. Also I can't use l'Hôpital or Taylor series. Any help would be appreciated, thanks in advance! (And sorry if I did anything wrong here)
 [/INST] First, you can decompose the first term in your derivative as:
$\frac{1}{x(1+x)} = \frac{1}{x}-\frac{1}{1+x}$.
Second, you can expand the second term using Taylor series:
$\frac{ln(1+x)}{x^2}$
$ = \frac{1}{x^2} \left( x - \frac{x^2}{2} + \frac{x^3}{3} - .. \right)$
$ = \frac{1}{x} - \frac{1}{2} + \frac{x}{3} - $
Putting it together, your derivative is:
$\frac{1}{x}-\frac{1}{1+x} - \frac{1}{x} + \frac{1}{2} - \frac{x}{3} +...$

*

*The $1/x$ terms cancel out.

*$\lim_{x \rightarrow 0} \frac{1}{1+x} = 1$.

*$\lim_{x \rightarrow 0}$ of all terms with $x$ or higher powers of $x$ in the numerator is 0.

So you are left with $-1 + \frac{1}{2} = -\frac{1}{2}$.
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