Value of $ \lim_{n \to \infty} \int \limits_{0}^{1}nx^n e^{ x^2} ?$ How to find the value of $$ \lim_{n \to \infty}\int \limits_{0}^{1} nx^n e^{ x^2} ?$$
From wolfram the limit approaches to $e$ for larger values of $n$. I substituted $x^2 $ with $u$ and obtained 
$$ \frac{ n} {2} \int \limits_{0}^{1} u^{\frac{n-1}{2}} e^{u} du $$ 
The value of this integral can be obtained from  here. But still I'm unable to get it. Is there any better approach for this question?

You can get an upper bound because for $0\le x\le1$, $e^{x^2}\le e$ so
$$\int_0^1nx^ne^{x^2}dx\le\frac{ne}{n+1}$$
We know that for $1-\epsilon\le x\le1$, $e^{x^2}\ge e^{1-2\epsilon+\epsilon^2}\ge e\cdot e^{-2\epsilon}\ge e(1-2\epsilon)$ because $e^{-x}\ge(1-x)$ for $x\ge0$, the latter function begin the linearization of the former at $x=0$ and the former being concave up. Also
$$1-(1-\epsilon)^{n+1}\ge1-e^{-\epsilon(n+1)}$$
So a lower bound is
$$\begin{align}\int_0^1nx^ne^{x^2}dx&\ge\int_{1-n^{-1/2}}^1nx^ne^{x^2}dx\ge\int_{1-n^{-1/2}}^1nx^ne(1-2n^{-1/2})dx\\
&=\frac{ne(1-2n^{-1/2})}{n+1}\left(1-(1-n^{-1/2})^{n+1}\right)\\
&\ge\frac{ne(1-2n^{-1/2})}{n+1}\left(1-e^{-n^{-1/2}(n+1)}\right)\end{align}$$
Since our integral lies between two quantities that approach $e$ as $n\rightarrow\infty$, the linit is in fact $e$.
