Prove $\lim_{x\rightarrow 0}\cos (x)=1$ with the epsilon-delta definition of limits Prove 
$$    
\lim_{x\rightarrow 0}\cos (x)=1
$$    
with the epsilon-delta definition of limits
 A: We want to  show that for every $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $0\lt |x-0|\lt \delta$, then $|1-\cos x|\lt \epsilon$.
To do this, we need an estimate of $|1-\cos x|$ for $x$ close to $0$. 
Suppose that $|x|\lt 1$.
Multiplying top and (missing) bottom by $1+\cos x$  (which is $\gt 1$ in our interval), we get 
$$|1-\cos x|=\frac{1-\cos^2 x}{1+\cos x}=\frac{\sin^2 x}{1+\cos x}.\tag{1}$$ 
In our interval, we have $\cos x\gt 0$, and therefore 
$$\frac{\sin^2 x}{1+\cos x}\le \sin^2 x.\tag{2}$$
It is always true that $|\sin x|\le x$. It follows from (1) and (2) that in our interval,
$$|1-\cos x|\le x^2\le x.\tag{3}$$
Finally, we have the required estimate. Let $\delta=\min(1,\epsilon)$. Making sure that $\delta\le 1$ is just to make sure that we are working in the interval $|x|\le 1$, since our inequality was derived under the assumption $|x|\le 1$. It is protection against someone giving us a ridiculous $\epsilon$, such as $\epsilon=10$.
If $|x|\lt \delta$, then by Inequality (3) we have  $|1-\cos x|\lt |x|\lt \delta=\epsilon$, so $|1-\cos x|\lt \epsilon$.  
In fact we have the stronger inequality $|1-\cos x|\le x^2$. So we could have chosen $\delta$ to be the minimum of $1$ and $\sqrt{\epsilon}$. 
Remark: An approach similar to the one in the answer may be intended. There are, however, issues, since the inequality was obtained by using "informal" geometric facts. For a fully formal approach, we should have a formal definition of $\cos x$, say via the power series. But then the result follows from general facts about power series.
A: Note that $0\leq 1-\cos x=2\sin^2\frac{x}{2}\leq2|\sin\frac{x}{2}|\leq|x|$, then:
$\forall\epsilon>0,\exists\delta=\epsilon>0$, s.t. $\forall x, |x|<\delta, |1-\cos x|\left(\leq|x|<\delta\right)<\epsilon$.
