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Worked Examples · Example 4

Q.Evaluate:

(i) lim⁡x→0sin⁡4xsin⁡2x\lim_{x\to 0}\dfrac{\sin 4x}{\sin 2x}
(ii) lim⁡x→0tan⁡xx\lim_{x\to 0}\dfrac{\tan x}{x}
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Both parts rest on the standard limit lim⁡θ→0sin⁡θθ=1\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta}=1. (i) equals 22;

(ii) equals 11.

The core idea

Near x=0x=0, both sin⁡x\sin x and tan⁡x\tan x behave like xx itself. The fundamental limit that makes this precise is:

lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1

From it we also get lim⁡θ→0tan⁡θθ=1\lim_{\theta \to 0} \dfrac{\tan \theta}{\theta} = 1, since tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta} and cos⁡θ→1\cos\theta \to 1.


(i) lim⁡x→0sin⁡4xsin⁡2x\displaystyle \lim_{x\to 0} \frac{\sin 4x}{\sin 2x}

Step 1 — Match each sine with its own argument. Multiply and divide so each sine sits over its own angle:

sin⁡4xsin⁡2x=sin⁡4x4x⋅2xsin⁡2x⋅4x2x=sin⁡4x4x⋅2xsin⁡2x⋅2\frac{\sin 4x}{\sin 2x} = \frac{\sin 4x}{4x}\cdot\frac{2x}{\sin 2x}\cdot\frac{4x}{2x} = \frac{\sin 4x}{4x}\cdot\frac{2x}{\sin 2x}\cdot 2

Step 2 — Take the limit of each factor. As x→0x\to 0, both 4x→04x\to 0 and 2x→02x\to 0, so

lim⁡x→0sin⁡4x4x=1,lim⁡x→02xsin⁡2x=1.\lim_{x\to 0}\frac{\sin 4x}{4x}=1,\qquad \lim_{x\to 0}\frac{2x}{\sin 2x}=1.

Step 3 — Combine. The constant factor 22 remains:

lim⁡x→0sin⁡4xsin⁡2x=1⋅1⋅2=2\lim_{x\to 0} \frac{\sin 4x}{\sin 2x} = 1\cdot 1\cdot 2 = 2 …

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