What is the result for this limit $\lim _{x\to \infty }\left(\frac{\left(\int _0^xe^{t^2}dt\:\right)^2}{\int _0^xe^{2t^2}dt\:}\right)$?

Question

what is the result for this limit? $$\lim _{x\to \infty }\left(\frac{\left(\int _0^xe^{t^2}dt\:\right)^2}{\int _0^xe^{2t^2}dt\:}\right)$$

Answer 1

Let $$ f(x)=\int_0^x e^{t^2}\,dt \qquad g(x)=\int_0^x e^{2t^2}\,dt $$ Then $$ f'(x)=e^{x^2} \qquad g'(x)=e^{2x^2} $$ Also, the derivative of $h(x)=(f(x))^2$ is $h'(x)=2f'(x)f(x)$.

Now you can apply l'Hôpital (twice, remember to simplify after the first step).