Find $\lim_{n\to \infty} \int_0^1 nx^n e^{x^2} dx$ The value of
$$\lim_{n\to \infty} \int_0^1 nx^n e^{x^2} dx$$
is ____________(round off to three decimal places)
I tried integrating by parts and bring out some recurrence relation , but it was of no use.
However, I can show the limit to be greater than or equal to 1.
$\int_0^1 nx^n e^{x^2} dx \gt \int_0^1 nx^{n+2} dx$ (using  $e^x\gt x $ for $x\gt 0$)
$=\frac n{n+3}=1-\frac 3{n+3}$
Taking limit as $n\to \infty$ , proves my assertion.
Please help me find the actual limit.Thanks in advance.
 A: Put $y=x^{n}$. The integral becomes $\int_0^{1} y^{1/n}e^{y^{2/n}}dy$. Check that the integrand is bounded by $e$ and that it converges point-wise to $e$. By DCT the answer is $e$. 
A: Alternatively,
\begin{equation}
\int_0^1 nx^n e^{x^2} dx = \int_0^1nx^n\sum_{k=0}^\infty \frac{x^{2k}}{k!} = \sum_{k=0}^\infty \int_0^1\frac{nx^{2k+n}}{k!} = \sum_{k=0}^\infty \frac{n}{(2k+n+1)k!}.
\end{equation}
Now,
\begin{equation}
\lim_{n\to\infty}\int_0^1 nx^n e^{x^2} dx = \sum_{k=0}^\infty\lim_{n\to\infty}\frac{n}{(2k+n+1)k!}= \sum_{k=0}^\infty\frac{1}{k!}=e
\end{equation}
A: $$I(n)=\int_0^1nx^ne^{x^2}dx$$
$u=x^2$ then $dx=\frac{du}{2x}$ and so:
$$I(n)=\frac n2\int_0^1x^{n-1}e^{x^2}du=\frac n2\int_0^1u^{\frac{n-1}{2}}e^udu$$
now:
$$e^u=\sum_{k=0}^\infty\frac{u^k}{k!}$$
and so:
$$I(n)=\frac{n}{2}\int_0^1\sum_{k=0}^\infty\frac{u^{\frac{2k+n-1}{2}}}{k!}du=n\sum_{k=0}^\infty\frac{1}{(2k+n-1)k!}$$
now notice that:
$$\lim_{n\to\infty}I(n)=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{n}{(2k+n-1)k!}=\sum_{k=0}^\infty\frac{1}{k!}=e$$
A: Late answer because the question might be tagged as duplicate where I put this answer before:
Partial integration gives
$$I_n :=\int_0^1 \underbrace{nx^{n-1}}_{u'}\cdot\underbrace{xe^{x^2}}_{v}dx= \left.x^{n+1}e^{x^2}\right|_0^1- \underbrace{\int_0^1 x^n(1+2x^2)e^{x^2}dx}_{J_n=}=e-J_n$$
Now, $J_n$ can be easily estimated as follows
$$0\leq J_n \leq 3e\int_0^1x^ndx=\frac{3e}{n+1}\stackrel{n\to \infty}{\longrightarrow}0$$
Hence, $I_n \stackrel{n\to \infty}{\longrightarrow} e$.
