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.
A: If $-1< c < 0$ then $\frac {1} {x^c(1-e^x)} \rightarrow -\infty$ as $x \downarrow 0.$ If $c=-1$ then  $\frac {1} {x^c(1-e^x)} \rightarrow -1$ as $x \downarrow 0.$ If $c<-1$ then  $\frac {1} {x^c(1-e^x)} \rightarrow 0$ as $x \downarrow 0.$
For $c=-1$ Kavi Rama Murthy sir has already explained below. For $-1<c<0$ write $d=-c.$ Then $0<d<1.$ Therefore $\frac {1} {x^c(1-e^x)} = \frac {x^d} {1-e^x}$ which is a $\frac 0 0$ form as $x \downarrow 0.$ Applying L'Hospital rule we get $\lim\limits_{x \downarrow 0} \frac {dx^{d-1}} {-e^{x}}.$ Since $d-1 < 0$ so $x^{d-1} \rightarrow \infty$ and $-e^{x} \rightarrow -1,$ as $x \downarrow 0.$ Hence $$\lim\limits_{x \downarrow 0} \frac {dx^{d-1}} {-e^{x}} = -\infty$$ since $d>0.$
Can you do it similarly for $c<-1$?
