Question

Find the limits $\underset{x\to 0^{+}}{lim}\frac{{\int }_0^{x^2}\sin \sqrt{x}dx}{x^3}$

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. Does not exist
  
• D. None of these
  

Answer 1

The correct option is A $\frac{2}{3}$

Given, $\frac{lim}{x\to 0^{+}}\frac{{\int }_0^{x^2}\sin \sqrt{x}dx}{x^3}$

$=\frac{lim}{x\to 0^{+}}\frac{2x\sin x}{{3x}^2}$

$=\frac{lim}{x\to 0^{+}}\frac{2\sin x}{3x}$

$=\frac{2}{3.}$