Q80 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
$\lim _{x \rightarrow 0} \dfrac{\sqrt{1-\cos 2 x}}{x \sqrt{2}}$
- Ais equal to -1
- Bis equal to 1
- Cis equal to 0
- Correct answerDdoes not exist
Solution
The given limit is $L = \lim _{x \rightarrow 0} \dfrac{\sqrt{1-\cos 2 x}}{x \sqrt{2}}$.
Using the trigonometric identity $1 - \cos 2x = 2 \sin^2 x$, we have $\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$.
Substituting this into the expression, $L = \lim _{x \rightarrow 0} \dfrac{\sqrt{2} |\sin x|}{x \sqrt{2}} = \lim _{x \rightarrow 0} \dfrac{|\sin x|}{x}$.
Evaluating the left-hand limit as $x \rightarrow 0^{-}$: $L^{-} = \lim _{x \rightarrow 0^{-}} \dfrac{-\sin x}{x} = -1$.
Evaluating the right-hand limit as $x \rightarrow 0^{+}$: $L^{+} = \lim _{x \rightarrow 0^{+}} \dfrac{\sin x}{x} = 1$.
Since the left-hand limit is not equal to the right-hand limit, the limit does not exist.
Answer: does not exist
Using the trigonometric identity $1 - \cos 2x = 2 \sin^2 x$, we have $\sqrt{1 - \cos 2x} = \sqrt{2 \sin^2 x} = \sqrt{2} |\sin x|$.
Substituting this into the expression, $L = \lim _{x \rightarrow 0} \dfrac{\sqrt{2} |\sin x|}{x \sqrt{2}} = \lim _{x \rightarrow 0} \dfrac{|\sin x|}{x}$.
Evaluating the left-hand limit as $x \rightarrow 0^{-}$: $L^{-} = \lim _{x \rightarrow 0^{-}} \dfrac{-\sin x}{x} = -1$.
Evaluating the right-hand limit as $x \rightarrow 0^{+}$: $L^{+} = \lim _{x \rightarrow 0^{+}} \dfrac{\sin x}{x} = 1$.
Since the left-hand limit is not equal to the right-hand limit, the limit does not exist.
Answer: does not exist
Hence, the correct answer is option: 'D'.