Calculate $\lim _{x \to \infty} (\int_0^x e^{t^2} dt)^{2}(\int_0^x e^{2 t^2} dt)^{-1}$

Question

Please find a mistake in my solution.

$$\lim _{x \to \infty} \frac{\left(\int_0^x e^{t^2} dt\right)^{2}} {\int_0^x e^{2 t^2} dt} = \text{|by d'Hospitals|} = \lim \limits_{x \to \infty} \frac{2 e^{x^2} \int_0^x e^{t^2} dt }{e^{2 x^2}} = 2 \lim_{x \to \infty} \frac{e^{x^2}}{2x e^{x^2}} = 0.$$

Unfortunately, Wolfram Alpha says that the solution is $\infty$.

Answer 1

Your answer is correct. Check your Wolfram Alpha code.