Question:
$\displaystyle\lim_{x \to 0} \frac{ \sqrt{1 -\cos \, 2x}}{\sqrt{2} x}$ is
$\displaystyle\lim_{x \to 0} \frac{ \sqrt{1 -\cos \, 2x}}{\sqrt{2} x}$ is
Updated On: Jul 5, 2022
- 1
- -1
- 0
- does not exist
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The Correct Option is D
Solution and Explanation
$\lim\frac{\sqrt{1-\cos 2x}}{\sqrt{2} x} \Rightarrow \lim \frac{\sqrt{1-\left(1-2 \sin^{2} x\right)}}{\sqrt{2} x} $
$\displaystyle\lim_{x \to0} \frac{\sqrt{2 \sin^{2} x}}{\sqrt{2} x} \Rightarrow \lim_{x \to0} \frac{\left|\sin x\right|}{x} $
The limit of above does not exist as
LHS = -1 $\neq $ RHL = 1
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Concepts Used:
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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.
Limits Formula:
Derivatives of a Function:
A 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.
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