Compute $\lim_{x \rightarrow +\infty} \frac{[\int^x_0 e^{y^{2}} dy]^2}{\int^x_0 e^{2y^{2}}dy}$ I've tried to apply L'hopitals rule on this one, as this get's $\frac{\infty}{\infty}$
$$\lim_{x \rightarrow +\infty} \frac{[\int^x_0 e^{y^2}\mathrm{d}y]^2}{\int^x_0 e^{2y^2}\mathrm{d}y}$$
$\frac{\mathrm{d} }{\mathrm{d} x}[\int^x_0 e^{y^2}\mathrm{d}y]^2 = 2[\int^x_0 e^{y^2}\mathrm{d}y] * [\frac{\mathrm{d} }{\mathrm{d} x}\int^x_0 e^{y^2}\mathrm{d}y]=2(e^{x^2}-1)(e^{x^2})$
and
$\frac{\mathrm{d} }{\mathrm{d} x}[\int^x_0 e^{2y^2}\mathrm{d}y]=e^{2x^2}$
so
$\lim_{x \rightarrow +\infty} \frac{[\int^x_0 e^{y^2}\mathrm{d}y]^2}{\int^x_0 e^{2y^2}\mathrm{d}y} = \lim_{x \rightarrow +\infty} \frac{2(e^{x^2}-1)(e^{x^2})}{e^{2x^2}} = 2\lim_{x \rightarrow +\infty} \frac{(e^{x^2}-1)}{e^{x^2}}=2\lim_{x \rightarrow +\infty} (1-\frac{1}{e^{x^2}})=2$
But the answer is $0$, so I think I've done a mistake somewhere I can't figure out where.
 A: You rather have
$$\frac{\mathrm{d} }{\mathrm{d} x}\left(\int^x_0 e^{y^2}\mathrm{d}y\right)^2 = 2\left(\int^x_0 e^{y^2}\mathrm{d}y\right) * \left(\frac{\mathrm{d} }{\mathrm{d} x}\int^x_0 e^{y^2}\mathrm{d}y\right)=2\color{red}{\left(\int^x_0 e^{y^2}\mathrm{d}y\right)}(e^{x^2})$$ giving 
$$
\lim_{x \rightarrow +\infty} \frac{2\color{red}{\left(\int^x_0 e^{y^2}\mathrm{d}y\right)}}{e^{x^2}}=\lim_{x \rightarrow +\infty} \frac{2e^{x^2}}{2xe^{x^2}}=\lim_{x \rightarrow +\infty}\frac1x=0.
$$
A: When you differentiate the numerator you should left with an integral and you need to use L'Hopital's rule twice. 
A: Without L'Hospital, we can use the estimates
$$
\begin{align}
\int_0^xe^{2t^2}\mathrm{d}t
&\ge\int_0^x\frac tx\,e^{2t^2}\mathrm{d}t\tag*{$1\ge\frac tx$}\\
&=\frac1{4x}\left(e^{2x^2}-1\right)
\end{align}
$$
and
$$
\begin{align}
\int_0^xe^{t^2}\mathrm{d}t
&\le\int_0^xe^{xt}\,\mathrm{d}t\tag*{$t\le x$}\\
&=\frac1x\left(e^{x^2}-1\right)
\end{align}
$$
to get
$$
\begin{align}
\frac{\left(\int_0^xe^{t^2}\mathrm{d}t\right)^2}{\int_0^xe^{2t^2}\mathrm{d}t}
&\le\frac4x\frac{\left(e^{x^2}-1\right)^2}{e^{2x^2}-1}\\
&=\frac4x\frac{e^{x^2}-1}{e^{x^2}+1}\\[3pt]
&\le\frac4x
\end{align}
$$
