Proof of Trig Limits — Question 7

Question 7

Prove that $$\lim_{x\to 0}\frac{\sin x}{\tan x}=1.$$

Question 7 - Solution

Rewrite the expression using the identity $$\tan x=\frac{\sin x}{\cos x}.$$

Substitute into the fraction: $$\frac{\sin x}{\tan x} = \frac{\sin x}{\dfrac{\sin x}{\cos x}} = \cos x,$$ for all $x$ where $\tan x$ is defined and $\sin x\neq 0$.

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

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

Therefore, $$\lim_{x\to 0}\frac{\sin x}{\tan x}=1.$$

$$\boxed{1}$$