Solve $\lim_{x \to \infty}e^{-x^2}\int_{x}^{x+\frac1x}e^{t^2}dt$ The solutions manual says
$$\lim_{x \to \infty}e^{-x^2}\int_{x}^{x+\frac1x}e^{t^2}dt=\lim_{x \to \infty}\frac{e^{(x+\frac1x)^2}-e^{x^2}}{2xe^{x^2}}$$
I'm trying to understand how they arrived there. Using L'Hôpital's rule rule, I have
$$\lim_{x \to \infty}e^{-x^2}\int_{x}^{x+\frac1x}e^{t^2}dt=
\lim_{x \to \infty}\frac{\int_{0}^{x+\frac1x}e^{t^2}dt-\int_{0}^{x}e^{t^2}dt}{e^{x^2}}=
\lim_{x \to \infty}\frac{e^{(x+\frac1x)^2}(1-\frac1{x^2})-e^{x^2}}{2xe^{x^2}}$$
I'm getting this additional factor $(1-\frac1{x^2})$ in $\frac{e^{(x+\frac1x)^2}(1-\frac1{x^2})-e^{x^2}}{e^{x^2}2x}$, which is $\frac{d}{dx}(x+\frac1x)$ and it comes from the application of the chain rule.
Is the chain rule not applicable there? My thinking was: let $F(x)=\int_{0}^{x}e^{t^2}dt$, and $G(x)=x+\frac1x$. Then $\int_{0}^{x+\frac1x}e^{t^2}dt=F(G(x))$, which is the functions' composition and I must apply the chain rule.
 A: Yes, the additional factor $(1-1/x^2)$ belongs there if you're doing L'Hopital. (Note, whether it's there or not, the limit turns out to be $0$, so the solution manual's assertion isn't technically wrong, it could claim to just be skipping a step, but it's a rather big step to skip. It's more likely an oversight on the part of the manual.)
As an alternative, without using L'Hopital, we can use the fact that $e^{t^2}$ is strictly increasing for $t\ge0$ to obtain the inequalities
$${e^{x^2}\over x}\le\int_x^{x+1/x}e^{t^2}\,dt\le{e^{(x+1/x)^2}\over x}={e^{x^2+2+1/x^2}\over x}$$
so that
$${1\over x}\le e^{-x^2}\int_x^{x+1/x}e^{t^2}\,dt\le{e^{2+1/x^2}\over x}$$
and now use the Squeeze Theorem to obtain
$$\lim_{x\to\infty}e^{-x^2}\int_x^{x+1/x}e^{t^2}\,dt=0$$
A: Hint-- use Leibniz rule of integration.

What they have done is basically use l's hospital rule. To take the derivative they have used the leibniz rule, and approximated $1/x^2$ to 0...
A: MVT for integrals as an option:
$f(x) = e^{-x^2}e^{z^2}\int_{x} ^{x+1/x}dx=$
$e^{-x^2}e^{z^2}(1/x)$, where $z \in [x, x+1/x];$
$1/x \le f(x) \le e^{-x^2}e^{(x+1/x)^2}(1/x);$
Squeeze.
