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)
 A: 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 $-1 + \frac{1}{2} = -\frac{1}{2}$.
A: Your original function $f(x)=\frac{\ln(1+x)}{x}$ can be written using the taylor series expansion of $\ln(1+x)$ as follows:
$$\ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{k}}{k}=x-\frac{x^2}{2}+\frac{x^3}{3}-...$$
$$\therefore\frac{\ln(1+x)}{x}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{k-1}}{k}=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{k}}{k+1}=1-\frac{x}{2}+\frac{x^2}{3}-...$$
$$\frac{d}{dx}\Bigg(\frac{\ln(1+x)}{x}\Bigg)=\frac{d}{dx}\bigg(1-\frac{x}{2}+\frac{x^2}{3}-...\bigg)=-\frac{1}{2}+\frac{2x}{3}-\frac{3x^2}{4}+...$$
So,
$$\lim_{x\to0}\Bigg(\frac{d}{dx}\Bigg(\frac{\ln(1+x)}{x}\Bigg)\Bigg)=\lim_{x\to0}\bigg(-\frac{1}{2}+\frac{2x}{3}-\frac{3x^2}{4}+...\bigg)=\fbox{$-\frac{1}{2}$}$$
A: $\frac {1}{x(x+1)} - \frac {\ln(x+1)}{x^2} = \frac {d}{dx} \frac {\ln(x+1)}{x}$
At which point I am inclined to use a series
$\frac {d}{dx} (1 - \frac x2 + \cdots)$
as $x$ goes to $0$ equals $-\frac 12$
Without using a series, I have.
$\frac {d}{dx} \frac {\ln(x+1)}{x}$ evaluated at $x = 0$
is $\lim_\limits{h\to 0} \frac{ \ln(1+h) - 1}{h^2}$
A: Because $\frac{x}{1+x}<ln(1+x)<x$
So when $ x \rightarrow  0+, f(x) \rightarrow  1$
When $x\rightarrow0-$ is similar
