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?
Let
$$I_n=\int_0^1x^ne^{x^2}dx$$
$I_n$ is decreasing because:
$$I_n-I_{n+1}=\int_0^1x^n(1-x)e^{x^2}dx\geq 0$$
We want to find $\lim\limits_{n\to \infty} nI_n$. Integrating by parts, we can see that:
$$I_n+\frac{2}{n+1}I_{n+2}=\frac{e}{n+1}$$
Now, since $I_n\geq I_{n+2}$, we have:
$$\frac{e}{n+1}\leq I_n+\frac{2}{n+1}I_n\Rightarrow I_n \geq \frac{e}{n+3}$$
Also because $I_n\leq I_{n-2}$, we get
$$\frac{e}{n-1}=I_{n-2}+\frac{2}{n-1}I_n\geq I_n+\frac{2}{n-1}I_n\Rightarrow I_n \leq \frac{e}{n+1}$$
Chaining these inequalities together, we have:
$$\frac{e}{n+3}\leq I_n \leq \frac{e}{n+1}$$
or
$$\frac{n}{n+3}e\leq nI_n\leq \frac{n}{n+1}e$$
Squeezing, we conclude that:
$$\lim\limits_{n\to \infty} nI_n = e$$
You have the right idea about changing the variable, just a different change: $u=x^{n+1}$ $$ \begin{align} \lim_{n\to\infty}\int_0^1nx^ne^{x^2}\,\mathrm{d}x &=\lim_{n\to\infty}\int_0^1\frac{n}{n+1}e^{u^{\frac2{n+1}}}\,\mathrm{d}u\\ &=\int_0^11\cdot e^1\,\mathrm{d}u\\[6pt] &=e \end{align} $$ Note that $\frac{n}{n+1}e^{u^{\frac2{n+1}}}$ increases monotonically to $e$ for all $u\in(0,1]$, and uniformly on compact subsets, so we can use monotone convergence, dominated convergence, or uniform convergence (on each compact subset) to justify the exchange of limit and integral.
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 $e$.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With Laplace Method:
\begin{align} \lim_{n \to \infty}\int_{0}^{1}nx^{n}\expo{x^{2}}\dd x & = \lim_{n \to \infty}\bracks{n\int_{0}^{1} \pars{1 - x}^{n}\expo{\pars{1 - x}^{2}}\dd x} \\[5mm] & = \lim_{n \to \infty}\bracks{n\int_{0}^{1} \expo{n\ln\pars{1 - x}}\expo{\pars{1 - x}^{2}}\dd x} \\[5mm] & = \lim_{n \to \infty}\bracks{n\int_{0}^{\infty} \expo{-nx}\expo{\pars{1 - 0}^{2}}\dd x} = \bbx{\Large \expo{}} \end{align}
