$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{1-x}}$ is equal to

  • A
    $1$
  • B
    does not exist
  • C
    $\sqrt{\frac{2}{3}}$
  • D
    $\ln 2$

Explore More

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3} = $

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ is

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right]$

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