Proving $\lim_{x \to 1^{-}} (1-x)^{r+1} \int_{1}^{\infty} s^r x^s ds = \int_{0}^{\infty} s^r e^{-s}ds$ 
Show that
$$\lim_{x \to 1^{-}} (1-x)^{r+1} \int_{1}^{\infty} s^r x^s ds = \int_{0}^{\infty} s^r e^{-s}ds$$

I tried taking the limit inside the integral but that didn't work. I also used properties of gamma function, including the limit definition
$$ \Gamma (r+1) = \int_{0}^{\infty} s^r e^{-s} ds = \lim_{n \to \infty} \dfrac{n^{r+1} n!}{(r+1)(r+2)\ldots(r+n+1)} $$
and then tried to manipulate the limit in the RHS, but it didn't work.
Any help will be appreciated.
 A: For $0 < x < 1$, we have,
$\displaystyle I_{x} (r) = (1-x)^{r+1} \int_{1}^{\infty}s^r x^s ds$
Let $u = -s \log x$
$\displaystyle \implies I_{x} (r) = \left(\dfrac{1-x}{-\log x} \right)^{r+1}\int_{-\log x}^{\infty} u^r x^\frac{-u}{\log x} \ du$
$\displaystyle = \left(\dfrac{1-x}{-\log x} \right)^{r+1}\int_{-\log x}^{\infty}u^r \left(x^\frac{-1}{\log x}\right)^u du$
$\displaystyle = \left(\dfrac{1-x}{-\log x} \right)^{r+1}\int_{-\log x}^{\infty}u^r e^{-u} du$
Since $\displaystyle \lim_{x \to 1^-} \left(\dfrac{(1-x)}{- \log x} \right) = 1$, we get,
$\displaystyle \lim_{x \to 1^-} I_{x} (r) = \lim_{x \to 1^-} \left(\dfrac{(1-x)}{- \log x} \right)^{r+1} \times \lim_{x \to 1^-}\int_{-\log x}^{\infty}u^r e^{-u} du $
$\displaystyle = \int_{0}^{\infty} u^r e^{-u} du$
$\therefore \displaystyle \lim_{x \to 1^{-}} (1-x)^{r+1} \int_{1}^{\infty} s^r x^s ds = \int_{0}^{\infty} s^r e^{-s}ds = \Gamma (r+1) \ \ \square$
A: Assuming $0<x<1.$ Let $x=e^{-y}.$ Then: $s^rx^s=s^re^{-ys}.$
Letting $t=sy $ $$\begin{align}\int_1^\infty s^rx^s\,ds&=y^{-1}\int_{y}^{\infty}\left(\frac{t}{y}\right)^re^{-t}\,dt\\
&=y^{-(1+r)}\int_{y}^\infty t^re^{-s}\,dt\end{align}$$
As $x\to 1^{-},$ we have $y\to 0^+.$ So, you get that you want:
$$\lim_{y\to 0^+} \left(1-e^{-y}\right)^{r+1}y^{-(1+r)}\int_y^{\infty} t^re^{-t}\,dt.$$
So you need $$\lim_{y\to 0^+} \left(\frac{1-e^{-y}}{y}\right)^{r+1}=1.$$
This is true since since $\frac{1-e^{-y}}{y}\to 1.$
