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.

Yes since you get a 0/0 form you can apply L-hopitals rule. Also all the derivatives and algebra looks perfect to me.
