Wrong initial intuition about the limit of $\frac{\sin(x)}{x}$ as $x$ goes to 0 I'm following the MIT OpenCourseware course on calculus, and when proving the derivative of $\sin(x)$ the following assumption is needed
$\lim\limits_{x\to0}  \frac{\sin(x)}{x} = 1$ 
The proof for this is given and I can follow it. 
My question however relates to my initial intuition. Which is that the limit of $\sin(x)$ at 0 is defined: $\sin(0) = 0$. So Why is it wrong to just plug that in 
$ \frac{\sin(0)}{0} = 0$
I would have never looked further because this seems like a simple problem. What is wrong with my intuition here?
 A: You are making two fundamental mistakes:
First: after substituting $x=0$, you wind up dividing by $0$.
This is an undefined expression; 
so, you should suspect that things have gone awry.
In your example, an even worse thing happens: you wind up with $0\over0$; 
which (if it makes any sense) is even more "undefined" than, say, $1\over 0$. 
For instance $\lim\limits_{x\rightarrow 0}{2x\over x}=2$
and $\lim\limits_{x\rightarrow 0}{ x\over 2x}={1\over2}$. 
In fact a 0/0 form could possibly be anything.
Secondly, when taking limits, you should really avoid just plugging in the limit point into the expression. 
This only works in certain cases (when the  limit exists and the function is continuous at the limit point), 
and, more importantly,  hides the true nature of what a limit is: 
Informally, 
you write $$\lim_{x \rightarrow a} f(x)=L$$ if the values of $f(x)$ get closer and closer to $L$ as $x$ gets 
closer and closer to, but not equal to, $a$.
Note the phrase "not equal to" in the above. This is an important restriction and is the reason why just 
substituting the limit point into what you're taking the limit of can lead to trouble.
As an example of determining the limit of a 0/0 form,  let's try to find
$$
\tag{1} \lim_{x\rightarrow 1}{(x+1)(x-1)\over (x-1)}
$$
Formally substitutiing $x=1$ in the above gives $0/0$.
If, however, $x$ is close to one, but different from 1, then  substitution into the 
right hand side of (1) gives a well-defined number;   moreover, the $(x-1)$ terms will cancel and the final result will be just $x+1$.
Thus, if $x$ is close to 1, but different from 1, 
the expression ${(x+1)(x-1)\over (x-1)}$ is close to 2. 
As it turns out, the closer $x$ is to 1, but different from 1, the closer 
${(x+1)(x-1)\over (x-1)}$ is to 2; and thus
$\lim\limits_{x\rightarrow 1}{(x+1)(x-1)\over (x-1)}=2$.
Finding your limit ($\thinspace\lim\limits_{x\rightarrow0}{\sin x\over x} \thinspace$)
is a bit more complicated than the above. You can find the standard derivation of this limit, which utilizes the 
Squeeze Theorem
and simple geometry here (or at your course site, as you said).
A: You can just use L'Hôpital's rule. 
$$\begin{align*}
\frac{d}{dx} \sin(x) &=\cos(x) \\
\frac{d}{dx} x &= 1 \\
\lim_{x\rightarrow0} \frac{\sin(x)}{x}&=\lim_{x\rightarrow0}\frac{\cos(x)}{1}=1
\end{align*}$$
Edit: I think this rule is very helpful to evaluate limits involving indeterminate forms. Not as proof but as a sanity check of an intuitive answer.
