Proof of Trig Limits — Question 5

Question 5 - Solution

We begin by using a trigonometric identity.

Recall that $$1-\cos(2x)=2\sin^2 x.$$

Substitute this into the expression: $$\frac{1-\cos(2x)}{x^2} = \frac{2\sin^2 x}{x^2}.$$

Rewrite the fraction: $$\frac{2\sin^2 x}{x^2} = 2\left(\frac{\sin x}{x}\right)^2.$$

Now take limits as $x\to 0$.

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

Therefore, $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^2=1^2=1.$$

Multiply by $2$: $$\lim_{x\to 0}\frac{1-\cos(2x)}{x^2} = 2\cdot 1 = 2.$$

$$\boxed{2}$$