Advertisement Remove all ads

Lim X → 0 | Sin X | X - Mathematics

MCQ

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

Options

  • 1          

  • −1       

  • 0             

  • does not exist 

Advertisement Remove all ads

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).

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Q 41 | Page 81
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×