$\epsilon - \delta$ proof of $\lim_{x \to 0} \frac{1-\cos x}{x}=0$

Question

My goal is to make $ \left| \frac{1-\cos x}{x} \right|< \epsilon $ by finding bounds for the function. Now, we know that
$$ \left| \frac{1-\cos x}{x} \right|< \frac{2}{ \left| x \right| } $$ We want this to be smaller than $\epsilon$, hence $\frac{2}{ \left| x \right| }<\epsilon \implies \left| x \right|> \frac{2}{\epsilon} $ $$ $$ Now the proof begins from here:$\ $ Let any $\epsilon >0$ and choose $\delta>\frac{2}{\epsilon}$. Then whenever $0< \left| x \right|<\delta$ $\ \ $implies $$ \left| \frac{1-\cos x}{x} \right|\le \frac{2}{ \left| x \right| } $$ Now I am stuck here, I want this last inequality to be smaller than $\frac{2}{ \delta } $ which it is not since $ \left| x \right|<\delta$ $\ \ $. It would complete the proof if $\frac{2}{ \left| x \right| } <\frac{2}{ \delta }<\frac{2}{ \frac{2}{\epsilon} }<\epsilon $

Answer 1

hint

You can use the fact that

$$1-\cos(x)=2\sin^2(\frac x2)$$

and

$$|\sin(X)|\le |X|$$

Answer 2

There's nothing actually wrong with your proof, it's just that, as you have noted, it doesn't help you get where you want. You want an upper bound for $|x|$ , not a lower one.