Evaluate $\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt$ Evaluate $\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt$
My approach:
$$
\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt = \lim_{x\rightarrow \infty} \frac{\int_{0}^{x}e^{t^2}dt}{x^{-1}e^{x^2}}
$$
Both the numerator and denominator $\rightarrow \infty$ as $x\rightarrow \infty$. Apply L'Hopital's rule and FTC:
$$
\lim_{x\rightarrow \infty} x\int_{0}^{x}e^{t^2-x^2}dt = \lim_{x\rightarrow \infty} \dfrac{e^{x^2}}{2e^{x^2}-x^{-2}e^{x^2}}=\frac{1}{2}
$$
I am looking for a verification of my result. Thank you!
 A: The numerator after applying L'Hospital's Rule in the OP is incorrect as it is written. Note that $$\frac{d}{dx} \int_0^x e^{t^2}\,dt=e^{x^2}\ne e^{x^2}-1$$

Applying L'Hospital's Rule reveals that
$$\begin{align}
\lim_{x\to \infty }x\int_0^x e^{t^2-x^2}\,dt&=\lim_{x\to \infty }\frac{\int_0^x e^{t^2}\,dt}{\frac{e^{x^2}}x}\\\\
&=\lim_{x\to \infty }\frac{e^{x^2}}{2e^{x^2}-\frac{e^{x^2}}{x^2}}\\\\
&=\lim_{x\to \infty }\frac{1}{2-\frac{1}{x^2}}\\\\
&=\frac12
\end{align}$$
and we are done!

I thought it might be instructive to present an approach that does not rely on L'Hospital's Rule.  To that end, we now proceed.
Enforcing the substitution $t\mapsto \sqrt{x^2-t}$, we have
$$\begin{align}
\lim_{x\to\infty} x\int_0^x e^{t^2-x^2}\,dt&=\lim_{x\to\infty} x\int_0^{x^2} \frac{e^{-t}}{2\sqrt{x^2-t}}\,dt\\\\
&=\frac12 \lim_{x\to\infty}\int_0^{x^2} \frac{e^{-t}}{\sqrt{1-\frac{t}{x^2}}}\,dt\\\\
&=\frac12\int_0^\infty e^{-t}\,dt\\\\
&=\frac12
\end{align}$$
as expected!
A: \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} x^{-1} \mathrm{e}^{x^2} 
    &= \frac{\mathrm{d}}{\mathrm{d}x}(x^{-1})\mathrm{e}^{x^2} + x^{-1}  \frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{x^2}  \\
    &= -x^{-2}\mathrm{e}^{x^2} + x^{-1}2x\mathrm{e}^{x^2}  \text{,}
\end{align*}
which is not what you have in your denominator...
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\Large\left. a\right)}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{x \to \infty}\pars{x\int_{0}^{x}\expo{t^{2} - x^{2}}\dd t}} =
\lim_{x \to \infty}\bracks{x\int_{0}^{x}\expo{\pars{x - t}^{2} - x^{2}}\dd t}
\\[5mm] = &
\lim_{x \to \infty}\bracks{x\int_{0}^{x}\expo{-2tx}\expo{-t^{2}}
\dd t}
=
\lim_{x \to \infty}\pars{x\int_{0}^{\infty}\expo{-2tx}\dd t}
\\[5mm] = &\
\lim_{x \to \infty}\pars{x\,{1 \over 2x}} = \bbx{1 \over 2} \\ &
\end{align}
See Laplace Method.

$\ds{\Large\left. b\right)}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{x \to \infty}
\pars{x\int_{0}^{x}\expo{t^{2} - x^{2}}\dd t}}
\lim_{x \to \infty}
\braces{x\bracks{{1 \over 2}\,\root{\pi}\expo{-x^{2}}\,{\mrm{erf}\pars{\ic x} \over \ic}}}
\end{align}
where $\ds{\mrm{erf}}$ is an
Error Function which has the asymptotic behavior $\ds{\mrm{erf}\pars{\ic x} \sim
1 - {\expo{x^{2}} \over \root{\pi}\ic x}}$. Then,
\begin{align}
&\bbox[5px,#ffd]{\lim_{x \to \infty}
\pars{x\int_{0}^{x}\expo{t^{2} - x^{2}}\dd t}} =
\bbx{1 \over 2} \\ &
\end{align}
