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प्रश्न
\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]
विकल्प
1
−1
0
does not exist
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उत्तर
We have,
\[\left| \sin x \right| = \begin{cases}\sin x, & 0 \leq x \leq \frac{\pi}{2} \\ - \sin x, & - \frac{\pi}{2} \leq x < 0\end{cases}\]
Now,
\[\lim_{x \to 0^-} \frac{\left| \sin x \right|}{x} = \lim_{x \to 0} \frac{- \sin x}{x} = - \lim_{x \to 0} \frac{\sin x}{x} = - 1\]
\[\lim_{x \to 0^+} \frac{\left| \sin x \right|}{x} = \lim_{x \to 0} \frac{\sin x}{x} = 1\]
Clearly,
\[\lim_{x \to 0^-} \frac{\left| \sin x \right|}{x} \neq \lim_{x \to 0^+} \frac{\left| \sin x \right|}{x}\]
∴\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\] does not exist.
Hence, the correct answer is option (d).
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