Skip to content
NCERT Exemplar · Q64

Q.lim⁡x→0∣sin⁡x∣x\lim_{x \to 0} \dfrac{|\sin x|}{x} is
(A) 11
(B) −1-1
(C) does not exist
(D) None of these

Jharkhand JacMCQ· 1mImportance★★★★★est
91% · 159/175 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The limit depends on whether xx approaches zero from the left or right because of the absolute value; the one-sided limits differ, so the limit does not exist.

The absolute value introduces a directional dependence. When xx is near zero, sin⁡x\sin x has the same sign as xx itself (positive for positive xx, negative for negative xx). This means ∣sin⁡x∣|\sin x| behaves differently on either side of the origin, and we must check the left-hand and right-hand limits separately.

The standard result lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 is fundamental, but here the absolute value changes the game.

Step-by-step analysis

  1. Right-hand limit (x→0+x \to 0^+):

    When x>0x > 0 (approaching from the right), we have sin⁡x>0\sin x > 0 for small positive xx, so ∣sin⁡x∣=sin⁡x|\sin x| = \sin x.

lim⁡x→0+∣sin⁡x∣x=lim⁡x→0+sin⁡xx=1\lim_{x \to 0^+} \frac{|\sin x|}{x} = \lim_{x \to 0^+} \frac{\sin x}{x} = 1

  1. Left-hand limit (x→0−x \to 0^-):

    When x<0x < 0 (approaching from the left), we have sin⁡x<0\sin x < 0 for small negative xx, so ∣sin⁡x∣=−sin⁡x|\sin x| = -\sin x.

lim⁡x→0−∣sin⁡x∣x=lim⁡x→0−−sin⁡xx=−lim⁡x→0−sin⁡xx=−1\lim_{x \to 0^-} \frac{|\sin x|}{x} = \lim_{x \to 0^-} \frac{-\sin x}{x} = -\lim_{x \to 0^-} \frac{\sin x}{x} = -1

  1. Comparing the one-sided limits: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.