Question

Evaluate the integral
$\underset{x\to 0}{lim}\frac{{\int }_0^{x^2}\tan \sqrt{t}dt}{x^3}$

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. Does not exist
• D. $\frac{3}{4}$

Answer 1

The correct option is C Does not exist

${lim}_{x\to 0}\frac{{\int }_0^{x^2}\tan \sqrt{t}dt}{x^3}$ has the $\frac{0}{0}$ form and hence, we use L'Hospital's rule.

$\therefore {lim}_{x\to 0}\frac{{\int }_0^{x^2}\tan \sqrt{t}dt}{x^3}={lim}_{x\to 0}\frac{\tan \sqrt{x^2}-\tan 0}{3x^2}={lim}_{x\to 0}\frac{\tan x}{3x^2}$

Again using L'Hospital's rule, we have

${lim}_{x\to 0}\frac{\tan x}{3x^2}={lim}_{x\to 0}\frac{{\sec }^{2}x}{6x}$

which has form $\frac{1}{0}$ which does not exist.