Finding $\lim_{x \to \infty}\int_0^x{e^{-x^2+t^2}}\,dt$ If we aren’t able to solve the integral $\int e^{-x^2}\,dx$, then how is it possible to find the $\lim_{x \to \infty}\int_0^x{e^{-x^2+t^2}}\,dt$? This was given to me by my prof, and I asked him multiple times if it was able to be solved. He said yes, but I’m just not getting it. Any thoughts?
 A: Here is, what I would say, a more standard analysis way. Admittedly a bit longer than applying l'Hospital, but maybe more instructive of why the limit is zero.
Let $0<\delta<1$ be fixed (but arbitrary). Then
$$
\int_0^x e^{-x^2+t^2}\,dt=\int_0^{\delta x} e^{-x^2+t^2}\,dt+\int_{\delta x}^xe^{-x^2+t^2}\,dt
$$
The first integral is estimated by the worst value of the integrand (attained when $t=\delta x$) times the length of the interval,
$$
\int_0^{\delta x} e^{-x^2+t^2}\,dt\leq \delta x e^{x^2(\delta^2-1)}
$$
For the second one, we note that when $\delta x<t<x$, $1/t<1/(\delta x)$, and so
$$
\begin{aligned}
\int_{\delta x}^xe^{-x^2+t^2}\,dt&=e^{-x^2}\int_{\delta x}^{x}\frac{1}{t} te^{t^2}\,dt\\
&<e^{-x^2}\frac{1}{\delta x}\int_{\delta x}^xte^{t^2}\,dt\\
&=e^{-x^2}\frac{1}{\delta x}\Bigl[\frac{1}{2}e^{t^2}\Bigr]_{\delta x}^x\\
&=\frac{1}{2\delta x}\bigl(1-e^{x^2(\delta^2-1)}\bigr).
\end{aligned}
$$
All in all
$$
0\leq \int_0^x e^{-x^2+t^2}\,dt<\delta x e^{x^2(\delta^2-1)}+\frac{1}{2\delta x}\bigl(1-e^{x^2(\delta^2-1)}\bigr).
$$
The squeeze theorem on limits now gives

$$\lim_{x\to+\infty}\int_0^x e^{-x^2+t^2}\,dt=0.$$

A: Let $I(x)$ be the integral defined by  
$$I(x)=\int_0^xe^{t^2}\,dt$$
First, split the integral into the sum
$$I(x)=\int_0^1e^{t^2}\,dt+\int_1^xe^{t^2}\,dt \tag 1$$
Now, integrating by parts the second integral in $(1)$ with $u=t^{-1}$ and $v=\frac12e^{t^2}$ reveals 
$$I(x)=\int_0^1e^{t^2}\,dt+\left(\frac{e^{x^2}}{2x}-\frac{e}{2}\right)+\frac12\int_1^x\frac{e^{t^2}}{t^2}\,dt \tag 2$$ 
For $x\ge 1$, we have
$$|I(x)|\le \left(\frac{e^{x^2}}{2x}+\frac{e}{2}\right)+\frac12\frac{e^{x^2}}{x^2}\le \frac32 \frac{e^{x^2}}{x}$$
and therefore, 
$$\left|e^{-x^2}I(x)\right|\le \frac{3}{2x}\to 0\,\,\text{as}\,\,x\to \infty$$

NOTE:
We can develop an asymptotic expansion for "large $x$"  by using Cauchy's Integral Theorem to rewrite $I(x)$ as
$$\begin{align}
I(x)&=\int_0^xe^{t^2}\,dt\\\\
&=\int_0^{i\infty} e^{z^2}\,dz+\int_{i\infty}^xe^{z^2}\,dz\\\\
&=\frac{i\sqrt \pi}{2}+\int_{i\infty}^xe^{z^2}\,dz \tag 3\\\\
\end{align}$$
Next, we integrate by parts the integral on the right-hand side of $(3)$ with $u=z^{-1}$ and $v=\frac12 e^{z^2}$ and obtain
$$I(x)=\frac{i\sqrt \pi}{2}+\frac{e^{x^2}}{2x}+\frac12\int_{i\infty}^x\frac{e^{z^2}}{z^2}\,dz \tag 4$$ 
Integrating by parts the integral on the right-hand side of $(4)$ with $u=z^{-3}$ and $v=\frac12 e^{z^2}$ yields 
$$I(x)=\frac{i\sqrt \pi}{2}+\frac{e^{x^2}}{2x}+\frac{e^{x^2}}{2^2\,x^3}+\frac34\int_{i\infty}^x\frac{e^{z^2}}{z^4}\,dz\tag 5 $$
We proceed by continuing to integrate by parts and find that
$$I(x)=\frac{i\sqrt \pi}{2}+\frac{e^{x^2}}{2x}\left(1+\sum_{n=1}^{N}\frac{(2n-1)!!}{2^n\,x^{2n}}\right)+\frac{(2N+1)!!}{2^{N+1}}\int_{i\infty}^x\frac{e^{z^2}}{z^{2(N+1)}}\,dz \tag 6$$
Using $(2\ell-1)!!=\frac{(2\ell)!}{2^{\ell}\,\ell!}$, we can write $(6)$ as 
$$I(x)=\frac{i\sqrt \pi}{2}+\frac{e^{x^2}}{2x}\left(1+\sum_{n=1}^{N}\frac{(2n)!}{n!\,(2x)^{2n}}\right)+\frac{(2N+1)!}{2^{2N+1}\,N!}\int_{i\infty}^x\frac{e^{z^2}}{z^{2(N+1)}}\,dz \tag 7$$
Note that for real values of $x$, the left-hand side of $(7)$ is purely real.  Therefore, we have
$$\bbox[5px,border:2px solid #C0A000]{I(x)=\frac{e^{x^2}}{2x}\left(1+\sum_{n=1}^{N}\frac{(2n)!}{n!\,(2x)^{2n}}\right)+\frac{(2N+1)!}{2^{2N+1}\,N!}\,\,\text{Re}\left(\int_{i\infty}^x\frac{e^{z^2}}{z^{2(N+1)}}\,dz\right) } \tag 8$$
It is important to realize that the partial sums in $(8)$ diverge as $N\to \infty$ for every value of $x$.  While this seems at first glance to make the expansion useless, this is not the case.  We can show, that the integral of the right-hand side of $(8)$ satisfies 
$$\left|\int_{i\infty}^x\frac{e^{z^2}}{z^{2(N+1)}}\,dz\right|= O\left(\frac{e^{x^2}}{x^{2N+1}}\right)$$
Therefore, $(8)$ provides a perfectly suitable expansion for large $x$.  
A word of caution is that for fixed $x$, the accuracy of the expansion eventually worsens with increasing $N$.  However, for fixed $N$, the accuracy of the expansion improves with increasing $x$.
A: Note that
$$\int_{0}^{x}{e^{-x^2+t^2}}\,dt=e^{-x^2}\int_{0}^{x}e^{t^2}\,dt=\frac{\int_{0}^{x}e^{t^2}\,dt}{e^{x^2}}$$
Now, to solve the limit, use Fundamental Theorem of Calculus and L'Hospital's Rule:
$$\lim_{x\to\infty}\frac{\int_{0}^{x}e^{t^2}\,dt}{e^{x^2}}=\lim_{x\to\infty}\frac{e^{x^2}}{2xe^{x^2}}=0$$
A: More generally,
consider
$I(f)
=\lim_{x \to \infty}
\frac1{f(x)}\int_0^x f(t)dt
$
where $f(x) \to \infty$.
I'll try to find
a condition on $f$
such that
$I(f) = 0$.
By L'Hopital's rule,
$I(f)
=\lim_{x \to \infty}
\frac1{f(x)}\int_0^x f(t)dt
=\lim_{x \to \infty}
\frac{f(x)}{f'(x)}
$
if this limit exists.
Therefore,
if
$\lim_{x \to \infty}
\frac{f(x)}{f'(x)}
=0
$,
then
$I(f) = 0$.
If
$f(x) = e^{x^a}$
with
$a > 1$
(the OP's case is $a=2$),
then
$f'(x)
=ax^{a-1}e^{x^a}
$,
so
$\frac{f(x)}{f'(x)}
=\frac1{ax^{a-1}}
\to 0
$
as $x \to \infty$.
