Question

The value of $\underset{x\to \infty }{lim}\frac{{\left(\underset{0}{\overset{x}{\int }}{e}^{x^2}dx\right)}^2}{\underset{0}{\overset{x}{\int }}{e}^{2x^2}dx}$ is

• A. $1$
• B. $2$
• C. $0$
• D. Does not exist

Answer 1

The correct option is C $0$

Since $e^{x^2}>0,e^{2x^2}>0$ in $[0,x],$ where $x>0$
So, $\underset{0}{\overset{x}{\int }}{e}^{x^2}dx$ and $\underset{0}{\overset{x}{\int }}{e}^{2x^2}dx\to \infty asx\to \infty$
So the given limit, $L=\underset{x\to \infty }{lim}\frac{{\left(\underset{0}{\overset{x}{\int }}{e}^{x^2}\right)}^2}{\underset{0}{\overset{x}{\int }}{e}^{2x^2}}$ is of the form $\frac{\infty }{\infty }.$
Therefore, Using L' Hospital's Rule
$L=\underset{x\to \infty }{lim}\frac{2e^{x^2}\underset{0}{\overset{x}{\int }}{e}^{x^2}dx}{e^{2x^2}}=2\underset{x\to \infty }{lim}\frac{\underset{0}{\overset{x}{\int }}{e}^{x^2}dx}{e^{x^2}}$
Again applying L-Hospital's Rule
$=2\underset{x\to \infty }{lim}\frac{e^{x^2}}{2x{e}^{x^2}}=\underset{x\to \infty }{lim}\frac{1}{x}=0$