Does limit means replacing $x$ for a number? I don't understand limit so much. For example I see $\lim_{x \to -3}$. And I always just put $-3$ everywhere I see $x$. I feel like I'm doing something wrong, but it seems correct all the time.
 A: 
(I need to put this onto a T-shirt. :)
A: Substitution "works" many times; it works but not always: $$\lim_{x\to a} f(x) = f(a)\quad \text{${\bf only}$ when $f(x)$ is defined and continuous at $a$}$$
and this is why understanding the limit of a function as the limiting value (or lack of one) when $x$ is approaching a particular value: getting very very near that value, is crucial. That is, $$\lim_{x \to a} f(x) \not\equiv f(a) \qquad\qquad\tag{"$\not \equiv$"$\;$ here meaning "not identically"}$$
E.g., your "method" won't work for $\;\;\lim_{x\to -3} \dfrac{x^2 - 9}{x + 3}\;\;$ straight off.
Immediate substitution $f(-3)$ evaluates to $\dfrac 00$ which is indeterminate: More work is required. Other examples are given in the comments. 
When we seek to find the limit of a function $f(x)$ as $x \to a$, we are seeking the "limiting value" of $f(x)$ as the distance between $x$ and $a$ grows increasingly small. That value is not necessarily the value $f(a)$.
And understanding the "limit" as the "limiting value" or lack there of, of a function is crucial to understanding, e.g. that $\lim_{x \to +\infty} f(x)$ requires examining the behavior of $f(x)$ as $x$ gets arbitrarily (increasingly) large, where evaluating $f(\infty)$ to find the limit makes no sense and has no meaning.  
A: Let's go back to the $\epsilon,\delta$ definition of a limit.
Let a function $f$ be defined on an open interval containing the real number $c$. -This mean we aren't going to do silly things like take the square root of negative numbers etc. IN some interval around $c$.
Let $L$ be a real number -Self explanatory.
Then $\lim_{x\rightarrow c}f(x)=L$ if and only if -This is about to be an equivalent statement to what comes next.
For any real number $\epsilon>0$ -So we're allowing $\epsilon$ to be as absolutely small as we like, as long as it's still positive.
There exists a real number $\delta$ with the following property -Ready for this?
For all $x$ if $0<|x-c|<\delta$ then $|f(x)-L|<\epsilon$ -This means that if $|x-c|$ (think of this as the distance between $x$ and $c$) is less than our number $\delta$ then $f(x)$ is arbitrarily close to the real number $L$.
I'm not sure if this was very helpful but let me end by suggesting that you look up visual representations of the above as well as visuals of where a limit does not exist.
A: One of the most important limits is the definition of the derivative
$$
\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}
$$
and you cannot substitute $h=0$ in that fraction.
