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.
 A: Besides using L'Hospital's Rule,
By the definition of derivative,
$\displaystyle\lim \limits_{x\to0}\frac{\ln(1+x+x^2+...+x^n)}{nx}=\lim \limits_{x\to0} \frac{\ln(1+x+x^2+...+x^n)-\ln1}{n(x-0)}=\left.\frac{1}{n}\frac{d}{dx}(ln(1+x+x^2+...+x^n))\right|_{x=0}$
$=\displaystyle\left.\frac{1}{n}\frac{1+2x+...+nx^{n-1}}{1+x+x^2+...+x^n}\right|_{x=0}=\frac{1}{n}$
A: Yes since you get a 0/0 form you can apply L-hopitals rule. Also all the derivatives and algebra looks perfect to me.
A: Without using L'Hospital,
$$\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}x$$
$$=-\lim_{x\to0}x^n\cdot\lim_{x\to0}\frac{\ln(1-x^{n+1})}{-x^{n+1}}+\lim_{x\to0}\dfrac{\ln(1-x)}{-x}$$
Use $\lim_{y\to0}\dfrac{\ln(1+u)}u=1$
A: Hint: $$S=1+x+x^2+\cdots+x^n$$ expand with $x$ to get $$xS=x+x^2+\cdots+x^{n+1}$$
Now, subtract both equations and solve for $S$:
$$S-xS=1-x^{n+1}$$
$$S=\frac{1-x^{n+1}}{1-x}$$
A: 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 $x\to 0$. 
