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.

Besides using L'Hospital's Rule,
By the definition of derivative,
$\displaystyle\lim \limits_{x\to0}\frac{\ln(1+x+x^2+...+x^n)}{nx}=\lim \limits_{x\to0} \frac{\ln(1+x+x^2+...+x^n)-\ln1}{n(x-0)}=\left.\frac{1}{n}\frac{d}{dx}(ln(1+x+x^2+...+x^n))\right|_{x=0}$
$=\displaystyle\left.\frac{1}{n}\frac{1+2x+...+nx^{n-1}}{1+x+x^2+...+x^n}\right|_{x=0}=\frac{1}{n}$
