<s>[INST] Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt:
\begin{align*}
\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\
&= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\
&= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{n(1+x+\dots+x^n)} \\
&= \frac{1}{n}.
\end{align*}
Are my steps correct? Thanks.
 [/INST] It is well-known that $\ln (1+z)= z+o(z)$ for small $z$, so plugging in $z=x+x^2+\cdots+x^n$ yields
$$\frac{\ln(1+x+\cdots+x^n)}{x}=\frac{x+\cdots+x^n+o(x+\cdots+x^n)}{x}=\frac{x+o(x)}{x}\to 1$$
as $x\to 0$. 
</s>