Q82 / 170 · Manipal (MET) 2020 · Mathematics

$\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \sin \sqrt{t} d t}{x^3}$ is equal to

  • Correct answer
    A
    $\frac{2}{3}$
  • B
    $\frac{1}{3}$
  • C
    $0$
  • D
    $\infty$

Solution

$\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \sin \sqrt{t} d t}{x^3} \quad\left(\right.$ form $\left.\frac{0}{0}\right)$ $=\lim _{x \rightarrow 0} \frac{\sin \sqrt{x^2} \times 2 x}{3 x^2}$ (using L' Hospital's rule) $=\lim _{x \rightarrow 0} \frac{2}{3} \frac{\sin x}{x}=\frac{2}{3} \times 1=\frac{2}{3}$

Hence, the correct answer is option: 'A'.

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