Limit: $\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$ How can I find the following limit? 
$$\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$$
It's a limit of type $\displaystyle 1^{\infty}$ and if I note with $\displaystyle f(x)=\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)$ then I must do: 
$$\displaystyle \lim_{x \to 0}{(1+f(x))}^{1/x}=\lim_{x \to 0}[(1+f(x))^{\frac{1}{f(x)}}]^{\frac{f(x)}{x}}. $$
Is ok? Thanks :)
 A: Write the expression above as
$$\lim_{x \to 0}\left [ 1 + \sum_{k=1}^n \log{(1+k x)}\right ]^{1/x}$$
Note that $n$ remains finite, so we can Taylor expand the logs to get
$$\lim_{x \to 0}\left [ 1 +\sum_{k=1}^n k x\right]^{1/x} = \lim_{x \to 0}\left(1+\frac12 n (n+1) x\right)^{1/x} = e^{n(n+1)/2}$$
Note that the error term in the Taylor expansion is $O(x^2)$, which in this limit produces an additional term as
$$(1+n(n+1)/2 x + c x^2)^{1/x} \sim  \left(1+\frac12 n (n+1) x\right)^{1/x} \left ( 1+\frac{c x^2}{1+n (n+1) x/2}\right )^{1/x} \sim\left(1+\frac12 n (n+1) x\right)^{1/x} e^{c x}$$
The error is then
$$(1+n(n+1)/2 x + c x^2)^{1/x} - (1+n(n+1)/2 x )^{1/x} = \left(1+\frac12 n (n+1) x\right)^{1/x} (e^{c x}-1)$$
As $x \to 0$, the error $\sim c x$.
A: You can also argue $$\lim_{x\to 0}\left[1+\sum_{k=1}^n\log(1+kx)\right]^{\frac{1}{x}}=e^{\lim_{x\to0}\frac{1}{x}\log\left[1+\sum_{k=1}^n\log(1+kx))\right]}$$
and then use L'Hopital's Rule to compute the remaining limit. (To spell this out rigorously, you're using the fact that you can pass limits through continuous functions. In other words, $$\lim_{x\to a}f(g(x))=f(\lim_{x\to a} g(x))$$ when $f$ is continuous at $\lim_{x\to a}g(x)$.) In fact, the most common trick for dealing with indeterminate forms of the form $1^\infty$ is to rewrite $\lim f(x)$ as $\lim e^{\log f(x)}$ and then use the fact I just mentioned.
