Q:

Evaluate the following limit: $\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}$

I need to evaluate the following limit:
$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c}
$$
for different values of the constant $c$.
What I've tried thus far:
We have that
$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c} = \lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$
Now I know that $1 - e^x \to 0$ as $x\to 0$. If $c\geq 0$ then $x^c\to 0$ as $x\to 0$ so we have that 
$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)} = \infty
$$
What I'm not so sure about is what happens when $c<0$. If $c< 0$ then we can write $x^c$ as $x^{-\gamma}$ where $\gamma = \left|c\right|$. I know that $\lim_{x\downarrow 0}x^{-\gamma}\to \infty$, but I'm not sure whether $x^{-\gamma}\to \infty$ quicker than $1 - e^x\to 0$ as $x\to 0$. I need to know because right now I'm not sure what the term $x^c(1 - e^x)$ does as $x\downarrow 0$ when $c <0$.
Question: 
How do I evaluate 
$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$
when $c< 0$?

A:

$\frac x {1-e^{x}} \to -1$ as $x \to 0+$ by L'Hopital's Rule. Multiply numerator by $x$ and change $x^{c}$ in the denominator to $x^{c+1}$. That doesn't change the expression. Is the rest clear now?

A:

Write it as:
$$\lim_{x\to 0}\dfrac{1}{x^c(1 - e^x)}=\lim_{x\to 0}\dfrac{x^{-c}}{1 - e^x}\stackrel{L'H}=\lim_{x\to 0}\dfrac{-cx^{-c-1}}{-e^x}=\\
\lim_{x\to 0}\dfrac{c}{x^{c+1}}=\begin{cases}-1, c=-1\\ \ \ \ 0, c<-1\\
-\infty, -1<c<0\\
+\infty, c>0\\
\end{cases}$$
If $c=0$, it is $\pm \infty$ depending on the side of approach.

