How to calculate $\lim\limits_{x \rightarrow 0} \frac{1}{1-\cos x}$

Question

I first tried to change the $1$ in the denominator so I could use the theorem $\cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2}$ But if I do that, the ultimate result will be $$\frac{1}{2\sin\frac{x}{2}\sin\frac{x}{2}}$$

I think I'm heading at it the wrong way. Could someone push me on the right track?

Answer 1

More simply we have

$$\cos x \to1^- \implies 1-\cos x \to 0^+$$

therefore

$$\lim\limits_{x \rightarrow 0} \frac{1}{1-\cos x}=\lim\limits_{y \rightarrow 0^+} \frac{1}{y}$$

---

As an alternative

$$\frac{1}{1-\cos x}=\frac{x^2}{1-\cos^2 x}\frac{1+\cos x}{x^2} \to 2\cdot \frac 2{0^+}$$