Question

Assertion :$\underset{x\to 0}{lim}\frac{\sqrt{1-\cos 2x}}{x}$ does not exist. Reason: $|\sin x|=\{\begin{array}{cc}\sin x;& 0<x<\frac{\pi }{2}\\ -\sin x;& -\frac{\pi }{2}<x<0\end{array}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Let, $L=\underset{x\to 0}{lim}\frac{\sqrt{1-\cos 2x}}{x}=\underset{x\to 0}{lim}\frac{\sqrt{2{\sin }^{2}x}}{x}=\underset{x\to 0}{lim}\frac{\sqrt{2}|\sin x|}{x}$
Thus $LHL=\underset{x\to 0^{-}}{lim}\frac{-\sqrt{2}\sin x}{x}=-\sqrt{2}$
and $RHL=\underset{x\to 0^{+}}{lim}\frac{\sqrt{2}\sin x}{x}=\sqrt{2}$
Clearly $L.H.L\ne R.H.L\Rightarrow$ given limit does not exist.
Also reason is correct and explaining the Assertion.