Proof of Trig Limits — Question 3

Question 3

Prove that $$\lim_{x\to 0}\frac{\sin(3x)}{x}=3.$$

Question 3 - Solution

We begin by rewriting the expression so that the fundamental trigonometric limit appears.

Write $$\frac{\sin(3x)}{x} = \frac{\sin(3x)}{3x}\cdot 3.$$

Now take limits as $x\to 0$.

Since $3x\to 0$ as $x\to 0$, the fundamental trigonometric limit gives $$\lim_{x\to 0}\frac{\sin(3x)}{3x}=1.$$

Therefore, $$\lim_{x\to 0}\frac{\sin(3x)}{x} = \left(\lim_{x\to 0}\frac{\sin(3x)}{3x}\right)\cdot 3 = 1\cdot 3 = 3.$$

$$\boxed{3}$$