$\lim_{x \rightarrow 0} \frac{\sin(\pi \sin^2 x)}{x^2} = $

  • A
    $\pi^2$
  • B
    $3\pi$
  • C
    $2\pi$
  • D
    $\pi$

Explore More

Similar Questions

Evaluate the limit: $\lim_{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right)$

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^{\sin x}} - 1}}{{{b^{\sin x}} - 1}} = $

The value of $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ is:

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = $

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo