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?

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}
$.
