Question

Evaluate $\underset{x\to \infty }{lim}\frac{{\left({\int }_0^x{e}^{x^2}dx\right)}^2}{{\int }_0^x{e}^{2x^2}dx}$.

Answer 1

$\underset{x\to \infty }{lim}\frac{{\left({\int }_0^x{e}^{{2x}^2}dx\right)}^2}{{\int }_0^x{e}^{{2x}^2}dx}\frac{\infty }{\infty }$form

Using L-Hospital's rule

$\underset{x\to \infty }{lim}\frac{2\left({\int }_0^x{e}^{{2x}^2}dx\right)(e^{{2x}^2}.1-0)}{e^{{2x}^2}.1-0}$

$=\underset{x\to \infty }{lim}\frac{2{\int }_0^x{e}^{{2x}^2}dx}{e^{{2x}^2}}\frac{\infty }{\infty }$form

$=\underset{x\to \infty }{lim}\frac{2e^{{2x}^2}.1-1}{2x{e}^{{2x}^2}}=\underset{x\to \infty }{lim}\frac{1}{x}$

$=0$