find the limit of the following expression Let $f\left(x\right)=\left(x^{2}+1\right)e^{x}$.
Find the following:$$\lim_{n\to+\infty}n\int_{0}^{1}\left(f\left(\frac{x^{2}}{n}\right)-1\right)\,dx$$
I dont know if L'Hopital's Rule may be used and tried using Integral Mean Value Theorem, but I cannot solve it! Any help?
 A: Using the substitution $u = x/\sqrt{n}$ we write
$$n\int_0^1 \left(f\left(\frac{x^2}{n}\right) - 1\right)\, dx = n^{3/2}\int_0^{1/\sqrt{n}} (f(u^2) - 1)\, du.$$
Now 
\begin{align}\lim_{n\to \infty} n^{3/2}\int_0^{1/\sqrt{n}} (f(u^2) - 1)\, du &= \lim_{h \to 0^{+}} \frac{\int_0^h (f(u^2) - 1)\, du}{h^3} \quad (\text{letting $h = 1/\sqrt{n}$})\\
&= \lim_{h\to 0^{+}} \frac{f(h^2) - 1}{3h^2} \quad (\text{by L'hospital's rule})\\
&= \lim_{t\to 0^{+}} \frac{f(t) - f(0)}{3t} \quad (\text{letting $t = h^2$})\\
&= \frac{1}{3}f'(0).
\end{align}
So $$\lim_{n\to \infty} n\int_0^1 \left(f\left(\frac{x^2}{n}\right) - 1\right)\, dx = \frac{f'(0)}{3}.$$
A: Here is a general technique to deal with such problems where the integrand is explicitly known. The way I presented it is not considered mathematically rigorous because it makes no sense to integrate a set, but the idea is to show that the integrand must actually be some function in some class, and since we already knew that it was integrable we can proceed to find the class of possible values of the integral. I chose this presentation to make the intuition clear.
Let $[c] = \{ x : |x| \le |c| \}$.
As $n \to \infty$:
  Let $a > 0$ such that $\exp(\frac{x^{2}}{n}) \in 1+\frac{x^{2}}{n}+[a]\frac{x^{4}}{n^2}$ for any $x \in [0,1]$.
  $n\int_{0}^{1}\left((\frac{x^4}{n^2}+1)\exp(\frac{x^{2}}{n})-1\right) dx$
  $\in n\int_{0}^{1}\left((\frac{x^4}{n^2}+1)(1+\frac{x^{2}}{n}+[a]\frac{x^{4}}{n^2})-1\right) dx$
  $\subseteq n\int_{0}^{1}\left(\frac{1}{n}x^2+[b]\frac{1}{n^2}\right) dx$ for some $b > 0$ [because $x^4,x^6,x^8 \in [0,1]$ and $\frac{1}{n^3},\frac{1}{n^4} \in [1]\frac{1}{n^2}$]
  $\subseteq \int_{0}^{1}x^2\ dx+[b]\frac{1}{n} \to \frac{1}{3}$
Note that this relied on the first-order term (in $n$) in the integrand to be $O(\frac{1}{n})$, in the same way that kobe's answer used L'Hopital's rule only on an indeterminate form. Note also that this method works even for non-differentiable integrands that have asymptotic expansions.
A: Explicitly,
$$
\lim_{n\to\infty} n \int_0^1 \left[f\left(\frac{x^2}{n}\right) -1\right]\,dx 
= \lim_{n\to\infty} n \int_0^1 \left[\left(\frac{x^4}{n^2} + 1\right)e^{x^2/n} - 1\right]\,dx
$$
You can expand the exponential as a Taylor series:
\begin{align*}
n\int_0^1 \left[\left(\frac{x^4}{n^2} + 1\right)e^{x^2/n} - 1\right]\,dx
&= n\int_0^1 \Biggl[\frac{x^4}{n^2}\left\{1 + \frac{x^2}{n} + \frac{1}{2}\frac{x^4}{n^2} + \frac{1}{3!}\frac{x^6}{n^3}+\cdots\right\} \\
&\qquad\qquad+1\left\{\color{red}{1} + \color{green}{\frac{x^2}{n}} + \frac{1}{2}\frac{x^4}{n^2} + \frac{1}{3!}\frac{x^6}{n^3}+\cdots\right\} \\
&\qquad\qquad\color{red}{-1}\Biggr]\,dx\\
&= n\int_0^1\left[\color{green}{\frac{x^2}{n}} 
+ \frac{3}{2}\frac{x^4}{n^2}
+ \frac{7}{6}\frac{x^6}{n^3} + \cdots
\right],dx
\end{align*}
Everything in the integrand after the first term is a multiple of $\frac{1}{n^2}$, so even after multiplying by $n$, the integrals will vanish as $n\to\infty$.  So the limit is
$
    \int_0^1 x^2\,dx = \frac{1}{3}
$.
