How find this limit: $\displaystyle \lim_{n\to\infty} \int_0^1 (1+x)^{-n-1}e^{x^2}\ dx$ 
I wish to find: $$\lim_{n\to\infty} n\int_0^1 (1+x)^{-n-1}e^{x^2}\ dx\ \ (
> n=1,2,\cdots)$$

I maybe have to evaluate:
$$\int_0^1 (1+x)^{-n-1}e^{x^2}\ dx$$
But I can't, can someone give me some help?
 A: Let $u=nx$ then we have
$$n\int_0^1 (1+x)^{-(n+1)}e^{x^2}dx=\int_0^n(1+u/n)^{-(1+n)}e^{(u/n)^2}du\to\int_0^\infty e^{-u}du=1$$
A: We can partially evaluate the integral, by integrating by parts,
$$\begin{align}
\int_0^1 \frac{n}{(1+x)^{n+1}}e^{x^2}\,dx &= \left[-\frac{e^{x^2}}{(1+x)^n} \right]_0^1 + \int_0^1 \frac{2xe^{x^2}}{(1+x)^n}\,dx\\
&= 1 - \frac{e}{2^n} + \int_0^1 \frac{2xe^{x^2}}{(1+x)^n}\,dx
\end{align}$$
The last two terms converge to $0$, so the limit is $1$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\expo}[1]{{\rm e}^{#1}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}$
The main contribution comes from values of $x \gtrsim 0$. Then,
\begin{align}\color{#ff0000}{\large%
\lim_{n \to \infty}n\int_{0}^{1}\pars{1 + x}^{-n - 1}\expo{x^{2}}\,{\rm d}x}
&=
\lim_{n \to \infty}n\int_{0}^{1}\expo{-\pars{n + 1}\ln\pars{1 + x}}\,{\rm d}x
\\[3mm]&=
\lim_{n \to \infty}n\int_{0}^{1}\expo{-\pars{n + 1}x}\,{\rm d}x
=
\lim_{n \to \infty}{n \over n + 1}\int_{0}^{n + 1}\expo{-x}\,{\rm d}x
\\[3mm]&=
\int_{0}^{\infty}\expo{-x}\,{\rm d}x
=
\color{#ff0000}{\Large 1}
\end{align}
