Question

The value of $\underset{x\to 0}{lim}\left[\frac{{\int }_0^{x^2}\left(\sin \sqrt{t}\right)dt}{x^3}\right]$ is

• A. $1/2$
• B. $2/3$
• C. $1/3$
• D. Does not exist

Answer 1

Answer: (D)

Does not exist

$\underset{x\to 0}{lim}\frac{{\int }_0^{x^2}\sin \left(\sqrt{t}\right)dt}{x^3}\left(\frac{0}{0}form\right)$

Using L’Hospital’s rule

$=\underset{x\to 0}{lim}\frac{\left(\sin \sqrt{x^2}\right)\cdot 2x}{3x^2}$

$=\underset{x\to 0}{lim}\frac{\sin |x|\cdot 2}{3x}=$ Does not exist. ($\therefore \sqrt{x^2}=|x|$)

$\therefore \underset{x\to 0}{lim}\frac{\sin |x|\cdot 2}{3x}=\frac{2}{3}$

$\underset{x\to 0}{lim}\frac{\sin |x|\cdot 2}{3x}=\frac{-2}{3}$

$\therefore$ Limit does not exist.