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!
 A: You can think of $\lim\limits_{x\to 0}f(x)=L$ to mean that as you approach $f(x)$ from the left of $0$ and the right of $0$, the function looks as if it is going to be that value, however it does not necessarily have to be that value. See below.

Let this curve be $g(x)$. $\lim\limits_{x\to 2}g(x)$ exists (because if I look at $g(x)$ from the left of $2$ and the right of $2$ it looks like its going to equal about $1$) but it is not equal to $g(2)$, which is actually $1.5$.
Extra Example:
Say I gave you a discontinuous function, such as $f(x)=\frac{1}{x}$, and I asked you to find $\lim\limits_{x\to 0}\frac{1}{x}$. Even without the graph, you can probably tell that this function is continuous everywhere except for $x=0$. (An easy way to see this is if you recall that division by zero is undefined and see all the ways that the denominator can possibly be $0$). For $f(x)=\frac{1}{x}$, only $x=0$ would make $\frac{1}{x}$ undefined. 
First off, if $\lim_{x\to 0}\frac{1}{x}$ exists, then $\lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)$, which just means that if I approach $f(x)$ from the left of $0$ or the right of $0$, I should get the same answer.
Now $\lim\limits_{x\to 0^+}\frac{1}{x}$ can be thought of as as $x$ gets smaller and smaller, what are the values of $\frac{1}{x}$ going to approach? Well, $\frac{1}{10}=0.1<\frac{1}{1}=1<\frac{1}{0.1}=10<\frac{1}{0.01}=100<\ldots <\frac{1}{0.000001}=1000000$.
As the $x$ values get smaller and smaller from the right of $0$,  $f(x)$ is getting larger and larger. We write this as $\lim\limits_{x\to 0^+}f(x)= \infty$
Now we will check what happens from the left of $0$. We will do a similar procedure as above. $\frac{1}{-1}=-1>\frac{1}{-0.1}=-10>\frac{1}{-0.001}=-1000>\ldots>\frac{1}{-0.000001}=-1000000$. 
This time we see that as the $x$ values get close and closer to $0$ from the left, $f(x)$ is getting smaller and smaller. We write this as $\lim\limits_{x\to 0^-}f(x)=-\infty$.
We note that $\lim\limits_{x\to 0^-}f(x)\neq \lim\limits_{x\to 0^+}f(x)$, and hence the limit does not exist. If the right sided  and left sided limits had been equal, then you could have said that the limit itself was equal.
A: In the definition of a limit, the value of the function at the point the limit is being taken to is specifically left out. 
We say that $\lim_{x\rightarrow a}{f(x)}=L$ if and only if for every $\epsilon>0$ there exists a $\delta>0$ such that $0<|x-a|<\delta$ implies that $|f(x)-L|<\epsilon$.  Notice that we ensured that $|x-a|>0$, so that it cannot be possible that $x=a$.  As others have pointed out, this shows the sense in which the limit determines what a function acts like at a point based on the nearby points.
For your specific example, the idea is that as $x$ approaches 0, $\sin x$ acts like $x$; for this reason, the limit as $x\rightarrow 0$ is 1.
One way in which you can evaluate it rigorously is using l'hopital's rule: if $\lim_{x\rightarrow a}{f(x)}=\lim_{x\rightarrow a}{g(x)}=0$, then $\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}=\lim_{x\rightarrow a}{\frac{f'(x)}{g'(x)}}$, assuming that the RHS exists and the derivatives $f'(x)$ and $g'(x)$ exist near $x=a$.
Thus, we would find that $$\lim_{x\rightarrow 0}{\frac{\sin x}{x}}=\lim_{x\rightarrow 0}{\cos x}=\cos 0=1.$$  
A: L'Hopital's rule is an overkill here. If you already know that $\sin'x = \cos x$ (which you need to know to use L'Hopital), then simply by definition of derivatives you have
$$
\lim_{x \to 0} \frac{\sin x}{x} = \sin'(0) = \cos 0 = 1.
$$
If you don't know that $\sin'x = \cos x$, then things start to depend on the definition of $\sin x$. If you define $\sin x$ using the power series $\sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k + 1}}{(2k+1)!}$, then proving $\lim_{x \to 0}\frac{\sin x}{x} = 1$ is not harder than proving that the definition is correct at all, i.e. that the series converges everywhere. If you have a geometric definition of $\sin x$, then the proof will be (at least in part) geometric.
A: 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..
