Which of the following function$(s)$ not defined at $x = 0$ has/have a removable discontinuity at $x = 0$?

  • A
    $f(x) = \frac{1}{\ln |x|}$
  • B
    $f(x) = \cos \left( \frac{|\sin x|}{x} \right)$
  • C
    $f(x) = x \sin \left( \frac{\pi}{x} \right)$
  • D
    All of the above

Explore More

Similar Questions

If the function $f(x)=\begin{cases} \frac{1-\cos x}{x^{2}}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then the value of $k$ is

If $f(x) = \begin{cases} [x] + [-x], & x \neq 2 \\ \lambda, & x = 2 \end{cases}$ is continuous at $x = 2$,then $\lambda = $ (where $[.]$ denotes the greatest integer function).

Let $f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{cases}$. Then at $x=0$:

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

If a function $f$ defined by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{cases}$ is continuous at $x=0$,then $a=$

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