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Lim X → 0 | Sin X | X - Mathematics

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Question

\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]

Options

  • 1          

  • −1       

  • 0             

  • does not exist 

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Solution

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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Chapter 29: Limits - Exercise 29.13 [Page 81]

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RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.13 | Q 41 | Page 81

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