Question:

$\displaystyle\lim_{x\to0} \frac{\sin^{2}x}{\sqrt{2} - \sqrt{1+\cos x}} $ equals :

Updated On: July 22, 2025
  • $2 \sqrt{2}$
  • $4 \sqrt{2}$
  • $ \sqrt{2}$
  • 4
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

$\lim_{x\to0} \frac{\left(\frac{\sin^{2}x}{x^{2}}\right)\left(\sqrt{2} + \sqrt{1+\cos x}\right)}{\left(\frac{1-\cos x}{x^{2}}\right)}$
$ = \frac{\left(1\right)^{2} .\left(2\sqrt{2}\right)}{\frac{1}{2}} = 4\sqrt{2}$
Was this answer helpful?
0
0

Top Questions on limits and derivatives

View More Questions

Questions Asked in JEE Main exam

View More Questions

JEE Main Notification

Concepts Used:

Limits And Derivatives

Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.

Limit of a Function

Limits Formula:

Limits Formula
 Derivatives of a Function:

derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.

 Derivatives of a Function

Properties of Derivatives:

Properties of Derivatives

Read More: Limits and Derivatives