Skip to content
NCERT Exemplar · Q17

Q.Evaluate lim⁡x→01−cos⁡2xx2\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}.

Odisha ChseShort· 2mImportance★★★★★est
64% · 112/175 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 is 22. We rewrite 1−cos⁡2x1 - \cos 2x using the double-angle identity, then use the standard limit lim⁡t→01−cos⁡tt2=12\lim_{t \to 0} \frac{1 - \cos t}{t^2} = \frac12 (or equivalently lim⁡u→0sin⁡uu=1\lim_{u \to 0} \frac{\sin u}{u} = 1) to evaluate directly.

The core idea: when xx is tiny, cos⁡2x\cos 2x is close to 11, so the numerator 1−cos⁡2x1 - \cos 2x is small — but so is x2x^2. The question is how small, relative to x2x^2. That ratio is exactly what the limit captures.

A direct substitution x=0x = 0 gives 00\frac{0}{0}, an indeterminate form. So we need to manipulate the expression into a form where the behaviour near zero is clear.


  1. Use the double-angle identity. Recall: cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x. Then 1−cos⁡2x=1−(1−2sin⁡2x)=2sin⁡2x1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x. So the limit becomes

lim⁡x→02sin⁡2xx2.\lim_{x \to 0} \frac{2\sin^2 x}{x^2}.

  1. Factor out the constant.

lim⁡x→02sin⁡2xx2=2⋅lim⁡x→0(sin⁡xx)2.\lim_{x \to 0} \frac{2\sin^2 x}{x^2} = 2 \cdot \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2.

  1. Apply the standard limit. The fundamental limit lim⁡x→0sin⁡xx=1\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 is a cornerstone of calculus. It tells us that near zero, sin⁡x\sin x behaves like xx. Therefore,

lim⁡x→0(sin⁡xx)2=12=1.\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 = 1^2 = 1.

  1. Multiply back. 2⋅1=2.2 \cdot 1 = 2. …

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.