$\underset{x\to \mathrm{\infty }}{lim} \frac{{\int }_0^{2x} x{e}^{x^2}dx}{e^{4x^2}}$ equals

Solution:

We have,

$\underset{x\to \mathrm{\infty }}{lim} \frac{{\int }_0^{2x} x{e}^{x^2}dx}{e^{4x^2}}=\underset{x\to \mathrm{\infty }}{lim} \frac{{\int }_0^{2x} e^{x^2}d\left(x^2\right)}{2e^{4x^2}}=\underset{x\to \mathrm{\infty }}{lim} \frac{{\left[e^{x^2}\right]}_0^{2x}}{2e^{4x^2}}$

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{{\int }_0^{2x} x{e}^{x^2}dx}{e^{4x^2}}=\underset{x\to \mathrm{\infty }}{lim} \frac{e^{4x^2}-1}{2e^{4x^2}}=\underset{x\to \mathrm{\infty }}{lim} \left(\frac{1}{2}-\frac{1}{e^{4x^2}}\right)=\frac{1}{2}$