Integration for exponential Prove that $$\lim _{x \rightarrow \infty} e^{-x^{2}}\int_{0}^{x} e^{t^{2}}  d t=0$$
What is the main intuition to tackle this problem? I tried to utilize all the possible theorems, including Taylor's theorem and other related ones, but all of them turned out to be in vain. How could possibly tackle the initial idea for such a seemingly easy exercise?
 A: L'Hopital's rule gives a one-liner proof of the equality. If you are determined to avoid this powerful machinery, however, then we can still prove this relatively easily.
Consider $x > 0$. Then
\begin{align*}
0
\leq e^{-x^2} \int_{0}^{x} e^{t^2} \, \mathrm{d}t
= \int_{0}^{x} e^{-(x-t)(x+t)} \, \mathrm{d}t
\leq \int_{0}^{x} e^{-(x-t)x} \, \mathrm{d}t
= \frac{1 - e^{-x^2}}{x}.
\end{align*}
So by the squeezing lemma, the limit is zero as desired.
A: One can write that
$$ \int_0^xe^{t^2}dt\leqslant xe^{x^2} $$
but the upper bound is not $o(e^{x^2})$. To solve this issue, we need an upper bound of $t\mapsto e^{t^2}$ that will be a $o(xe^{x^2})$, for instance something like $e^{x^2-x^{\alpha}}$ with $\alpha\in]0,2[$. To do so, you can reduce the upper bound of the integral, but you need to make sure that the second part will still be negligeable, beside having an upper bound in the integral that is not $o(e^{x^2})$. To make sure this does not happen, it suffices to let the interval of integration going to $0$. We can summarize this as follow :
$$ \int_0^{x}e^{t^2}dt=\int_0^{x-\frac{1}{\sqrt{x}}}e^{t^2}dt+\int_{x-\frac{1}{\sqrt{x}}}^x e^{t^2}dt\leqslant \left(x-\frac{1}{\sqrt{x}}\right)e^{x^2-2\sqrt{x}+\frac{1}{x}}+\frac{1}{\sqrt{x}}e^{x^2} $$
Therefore
$$ e^{-x^2}\int_0^xe^{t^2}dt\leqslant e^{-2\sqrt{x}+\frac{1}{x}}+\frac{1}{\sqrt{x}}\underset{x\rightarrow +\infty}{\longrightarrow}0 $$
A: One liner with l'Hopital's
$$\lim _{x \rightarrow \infty} \frac{\int_{0}^{x} e^{t^{2}} d t}{e^{x^2}} \: \stackrel{\text{l'Hop.}}{=} \: \lim _{x \rightarrow \infty} \frac{e^{x^{2}} }{2xe^{x^2}}= \lim _{x \rightarrow \infty} \frac{1}{2x} =0$$
