$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2} = $

  • A
    $\frac{\pi}{2}$
  • B
    $\frac{\pi^2}{4}$
  • C
    $\frac{\pi^2}{2}$
  • D
    $\frac{\pi}{4}$

Explore More

Similar Questions

The value of $\lim_{x \rightarrow 0} \frac{|x|}{x}$ is

$\lim _{x \rightarrow \infty}\left[\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right]^{\frac{4 x+3}{8 x-1}} = $

$\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)(n^2+3 n-1)}$

The set of all values of $a$ for which $\lim_{x \rightarrow a}(\lfloor x-5 \rfloor - \lfloor 2x+2 \rfloor) = 0$,where $\lfloor \alpha \rfloor$ denotes the greatest integer less than or equal to $\alpha$,is equal to

If a function $f$ is defined by $f(x) = \frac{\cot^3 x - \tan x}{\cos(x + \pi/4)}$ for $x \neq \pi/4$,then $\lim_{x \rightarrow \pi/4} f(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