Proof of Trig Limits — Question 9

Question 9

Prove that $$\lim_{x\to 0}\frac{\sin(2x)}{\tan x}=2.$$

Question 9 - Solution

We begin by rewriting the expression using trigonometric identities.

Recall that $$\tan x=\frac{\sin x}{\cos x}.$$

Substitute this into the fraction: $$\frac{\sin(2x)}{\tan x} = \sin(2x)\cdot\frac{\cos x}{\sin x}.$$

Next, use the identity $$\sin(2x)=2\sin x\cos x.$$

Substitute: $$\frac{\sin(2x)}{\tan x} = (2\sin x\cos x)\cdot\frac{\cos x}{\sin x}.$$

Cancel $\sin x$ (for $x\neq 0$): $$\frac{\sin(2x)}{\tan x} = 2\cos^2 x.$$

Now take the limit as $x\to 0$.

Since cosine is continuous, $$\lim_{x\to 0}\cos x=\cos 0=1.$$

Therefore, $$\lim_{x\to 0}2\cos^2 x = 2(1)^2 = 2.$$

$$\boxed{2}$$