Proof of Trig Limits — Question 2

Question 2

Prove that $$\lim_{x\to 0}\frac{1-\cos x}{x}=0.$$

Question 2 - Solution

We begin by rewriting the expression using a trigonometric identity.

Recall the identity $$1-\cos x = 2\sin^2\!\left(\frac{x}{2}\right).$$

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

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

Now rewrite the second factor: $$\frac{2\sin\!\left(\frac{x}{2}\right)}{x} = \frac{\sin\!\left(\frac{x}{2}\right)}{\frac{x}{2}}.$$

Thus, $$\frac{1-\cos x}{x} = \sin\!\left(\frac{x}{2}\right) \cdot \frac{\sin\!\left(\frac{x}{2}\right)}{\frac{x}{2}}.$$

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

Since $\displaystyle \lim_{u\to 0}\sin u = 0$, we have $$\lim_{x\to 0}\sin\!\left(\frac{x}{2}\right)=0.$$

Also, using the fundamental trig limit $$\lim_{u\to 0}\frac{\sin u}{u}=1,$$ we obtain $$\lim_{x\to 0}\frac{\sin\!\left(\frac{x}{2}\right)}{\frac{x}{2}}=1.$$

Therefore, by the product law for limits, $$\lim_{x\to 0}\frac{1-\cos x}{x} = 0\cdot 1 = 0.$$

$$\boxed{0}$$