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NCERT Exemplar · Q18

Q.Evaluate lim⁡x→02sin⁡x−sin⁡2xx3\lim_{x \to 0} \dfrac{2\sin x - \sin 2x}{x^3}.

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Factor the numerator as 2sin⁡x−sin⁡2x=2sin⁡x (1−cos⁡x)2\sin x - \sin 2x = 2\sin x\,(1-\cos x), then use the standard limits sin⁡xx→1\dfrac{\sin x}{x}\to 1 and 1−cos⁡xx2→12\dfrac{1-\cos x}{x^2}\to \dfrac12. The value is 11.

Step 1 — Check the form

At x=0x=0: the numerator is 2sin⁡0−sin⁡0=02\sin 0 - \sin 0 = 0 and the denominator is 03=00^3 = 0, so we have 00\dfrac{0}{0} and must simplify.

Step 2 — Factor the numerator

Use the double-angle identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x:

2sin⁡x−sin⁡2x=2sin⁡x−2sin⁡xcos⁡x=2sin⁡x (1−cos⁡x).2\sin x - \sin 2x = 2\sin x - 2\sin x\cos x = 2\sin x\,(1 - \cos x).

Step 3 — Split into standard pieces

2sin⁡x−sin⁡2xx3=2sin⁡x (1−cos⁡x)x3=2⋅sin⁡xx⋅1−cos⁡xx2.\frac{2\sin x - \sin 2x}{x^3} = \frac{2\sin x\,(1-\cos x)}{x^3} = 2\cdot\frac{\sin x}{x}\cdot\frac{1-\cos x}{x^2}. …

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