How to solve this limit? $ \lim_{x \to +\infty} \frac{(\int^x_0e^{x^2}dx)^2}{\int_0^xe^{2x^2}dx}$ There's  a limit which confuses me:
$$ \lim_{x \to +\infty} \dfrac{(\int^x_0e^{x^2}dx)^2}{\int_0^xe^{2x^2}dx}$$
Is it possible to use L'Hôpital's rule here?

An asymptotic way:
Let $\displaystyle F:x\mapsto \int_0^x e^{t^2} dt$.
Note that $$\frac{(\int^x_0e^{t^2}dt)^2}{\int_0^xe^{2t^2}dt} = \sqrt 2\frac{(F(x))^2}{F(\sqrt 2 x)}$$
Moreover, $$\begin{align}
F(x)&=\int_0^1 e^{t^2} dt + \int_1^x \frac{1}{2t}\cdot 2te^{t^2}dt \\
&= \int_0^1 e^{t^2} dt + \frac{e^{x^2}}{2x}-\frac e2  +\frac 12 \int_1^x \frac{e^{t^2}}{t^2}dt
\end{align}$$
The same trick performed on $\displaystyle \int_1^x \frac{e^{t^2}}{t^2}dt$ yields
$$\begin{align}
F(x)&= \int_0^1 e^{t^2} dt + \frac{e^{x^2}}{2x}-\frac e2  + \frac{e^{x^2}}{4x^3} - \frac e4 + \frac 34 \int_1^x \frac{e^{t^2}}{t^4}dt
\end{align}$$
Note that for large enough $x$, $\displaystyle \int_1^x \frac{e^{t^2}}{t^4}dt\leq (x-1)\frac{e^{x^2}}{x^4}\leq \frac{e^{x^2}}{x^3}$.
Therefore, $$F(x)=\frac{e^{x^2}}{2x} + \frac{e^{x^2}}{4x^3} + O\left(\frac{e^{x^2}}{x^3}\right)$$
For the purpose of the question, it is enough to write $$F(x)=\frac{e^{x^2}}{2x} + o\left(\frac{e^{x^2}}{2x}\right)$$
which is the same as $\displaystyle F(x)\sim \frac{e^{x^2}}{2x}$. This implies $\displaystyle F(\sqrt 2 x)\sim \frac{e^{2x^2}}{2\sqrt 2x}$
Therefore, $$\frac{(\int^x_0e^{t^2}dt)^2}{\int_0^xe^{2t^2}dt} = \sqrt 2\frac{(F(x))^2}{F(\sqrt 2 x)}\sim \frac 1x\xrightarrow[x\to \infty]{}0$$
