application of L'Hopital's rule? I am trying to evaluate the following  limit: 
$$
\lim_{x \to 0} \frac{e^x}{\sum_{n = 1}^\infty n^k e^{-nx}},
$$
where $k$ is a large (but fixed) positive integer. I am unsure how to proceed. Can this be done using L'Hopital's rule? Just started learning calculus, thanks guys!!
 A: For each $x\lt\frac12$, there is an $n\in\mathbb{N}$ so that $1\lt\frac1x-1\le n\lt\frac1x$. Picking this one term out gives
$$
\sum_{n=0}^\infty n^ke^{-nx}\ge \left(\frac1x-1\right)^ke^{-1}
$$
As $x\to0^+$, $\left(\frac1x-1\right)^ke^{-1}\to\infty$
A: For any $N\in \mathbb N,$ we have for $x>0$ that
$$\frac{e^x}{\sum_{n=1}^{\infty}n^ke^{-nx}} < \frac{e^x}{N^ke^{-Nx}}.$$
Thus
$$0\le \limsup_{x\to 0^+} \frac{e^x}{\sum_{n=1}^{\infty}n^ke^{-nx}} \le \limsup_{x\to 0^+}\frac{e^x}{N^ke^{-nx}} = \lim_{x\to 0} \frac{e^x}{N^ke^{-Nx}} = \frac{1}{N^k}.$$
Since $N$ is arbitrary, the $\limsup$ on the left is arbitrarily small, hence equals $0.$ Thus the limit is $0$ from the right.
(Note that to the left of $0$ the sum in the denominator of our expression is identically $\infty.$ Not sure what to make of that.)
A: I think, the whole converges to 0 (zero), because the numerator converges to 1 and the denominator converges to infinity.
A: You don't need L'Hopital's rule; try substituting $0$ for $x$ and see what you get.
