About $ \lim_{x\rightarrow 0}\frac {\sin x}{x} = 1$ I do not understand how $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ As if $$ x = 0, \frac{\sin (0)}{0} = \frac {0}{0} $$ 
So if someone could explain this I would appreciate it!
Thanks!

Easy way to view: L'Hospital Rule: 
$$ \lim_{x\rightarrow 0} \frac {\sin x}{x} = \lim_ {x \rightarrow 0} \frac {\cos x}{1} = \cos 0 = 1 $$ 
Another way: first take $ 0 <\delta <\frac {\pi}{2} $. Thus, $ \forall x \in (0, \delta) $ we have the following chain of valid inequalities: 
$$ \sin x \leq x \leq \tan x $$ 
As $ 0 <x <\frac {\pi}{2} $, dividing this chain of inequalities for $ \sin x \neq $ 0, we obtain 
$$ 1 \leq \frac {x}{\sin x} \leq \frac {1}{\cos x}. $$ 
Taking the inverse, 
$$ \cos x \leq \frac {\sin x}{x} \leq  1 $$ 
Now, for the "sandwich theorem", making the limit as $ x \rightarrow 0^+ $, we conclude that 
$$ \lim_ {x \rightarrow 0^+} \frac {\sin x}{x} = 1. $$ 
Similarly it is shown that $$ \lim_{x \rightarrow 0^-} \frac{\sin x}{x} = 1. $$ 
For this, take $ x \in (- \delta, 0) $, in this case $ \sin x <$ 0, etc..
