Proof of Trig Limits — Question 4

Question 4 - Solution

We rewrite the expression in terms of sine and cosine.

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

Substitute this into the limit: $$\frac{\tan x}{x} = \frac{\sin x}{x\cos x} = \left(\frac{\sin x}{x}\right)\left(\frac{1}{\cos x}\right).$$

Now take limits of each factor as $x\to 0$.

From the fundamental trigonometric limit, $$\lim_{x\to 0}\frac{\sin x}{x}=1.$$

Also, since cosine is continuous at $0$, $$\lim_{x\to 0}\cos x=\cos 0=1,$$ and therefore $$\lim_{x\to 0}\frac{1}{\cos x}=1.$$

By the product law for limits, $$\lim_{x\to 0}\frac{\tan x}{x} = 1\cdot 1 = 1.$$

$$\boxed{1}$$