Question

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

• A. 0
• B. $\infty$
• C. 2
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

Consider $\underset{x\to \infty }{\lim }\frac{\underset{0}{\overset{2x}{\int }}x{e}^{x^2}dx}{e^{4x^2}}$
$=\underset{x\to \infty }{\lim }\frac{2\underset{0}{\overset{2x}{\int }}x{e}^{x^2}dx}{2e^{4x^2}}$
$=\underset{x\to \infty }{\lim }\frac{2\underset{0}{\overset{2x}{\int }}{e}^{x^2}d\left(x^2\right)}{2e^{4x^2}}$
$=\underset{x\to \infty }{\lim }\frac{{\left[e^{x^2}\right]}_0^{2x}}{e^{4x^2}}$
$=\underset{x\to \infty }{\lim }\frac{e^{x^2}-1}{2e^{4x^2}}$
$=\underset{x\to \infty }{\lim }\left(\frac{1}{2}-\frac{1}{2e^{4x^2}}\right)$
$=\frac{1}{2}$