Please clear up this misunderstanding I have involving limits. I have passed the calc sequence, and thought that I understood things.  But I have trouble with limits now and again.  From the way I understand it, the 'best' way to evaluate a limit, is to evaluate the expression at the number in the limit.  In that way it's not only a limit, but an algebraic expression (so, I think, the need for limits goes away entirely, superseded by accuracy.)
But then I see something like $\lim_{x\to 2} \frac{x^2-4}{x-2}$.  Now, I think in my mind 'the best way to evaluate this is just to plug in $2$'  But it doesn't work that way, as the limit is $4$, and if I plug in, the answer is $0/0$.  I think that people here, with their natural talent in math, can perceive where my train of thought is going wrong, and correct me.  Please!
 A: Notice that 
$$
  \lim_{x\to 2} \frac{x^2-4}{x-2} \;=\; 
  \lim_{x\to 2} \frac{(x-2)(x+2)}{x-2} \;=\; 
  \lim_{x\to 2} (x+2) \;=\; 
  4
$$
The idea is that $\frac{x^2-4}{x-2}$ has a point discontinuity at $x=2$: it has a little hole there because we can't divide by zero. Other than that little hole, the expressions $\frac{x^2-4}{x-2}$ and $(x+2)$ are the same. We can see in each of them that the limit as $x$ approaches $2$ is $4$.
Also, in response to what you said

From the way I understand it, the 'best' way to evaluate a limit, is to evaluate the expression at the number in the limit. In that way it's not only a limit, but an algebraic expression (so, I think, the need for limits goes away entirely, superseded by accuracy).

I see what you are saying, but I would phrase it a little differently. In the cases where you can just evaluate a $\lim_{x\to n}f(x)$ by plugging $n$ into $f(x)$, I would word it as "the limit of $f(x)$ as $x$ approaches $n$ happens to be the same as the value of $f$ at $n$".
A: Typically limits are used when the expression isn't defined where you want to evaluate it, so just plugging the limit in, will rarely do anything.
But in cases like the one you mention, you can rewrite the expression so something that is defined at the limit, and then you can plug in the limit.
What you need to observe is that $x^2-4 =(x-2)(x+2)$.
A: First of all: why do we take limits?
If you can just plug a number in an expression, why do we use $\epsilon-\delta$ to define what a limit is?
Take $f(x) = x^2$
Here the best way to evaluate a limit is just to plug the number in, as you say; this is because $$\lim_{x \to x_0} f(x) = f(x_0)$$ in every point $x_0$. Functions with this property are called continuous and you can just plug the numbers in to find the result (that is, calculate $f(x_0)$)
The problem is that sometimes the value $f(x_0)$ is not defined..
Take $f(x) = \frac {x^2}{x}$. What is $f(0)$? It's not defined.
But if you draw the function $f(x)$ , you can see that near $0$ the function is basically $0$. So even though it does not exists in $0$, in a certain sense for values of $x$ very close to $0$ the function is equal to $0$. 
The precise way of formulating this intuition is to define what a limit is! And you find out that $\lim_{x \to 0} \frac{x^2}{x} = 0$ as we wanted.

Now how do we find limit, in practice? 
As you said, if you just plug $2$ in your function it will be undefined. 
But when we calculate limit we only care about what happens near the point (in your case $2$) not at the point. 
Since your function is equal to $x+2$ for every point $x \neq 2$, you can just plug $2$ in this simpler expression to find out that the limit is $4$
A: The limit of a function $f$ at the point $a$ is not the value of $f$ at $a$. Those are two different things. It just so happens that for many functions, they are equal.
The limit of a function is the value that the function goes towards as you approach that point. That doesn't necessarily tell you anything about the value of the function at that point.
In your case, you have a function which has a certain limit as $x$ approaches $2$, yet which isn't even defined at $x=2$. This isn't paradoxical because nobody ever said that $\lim_{x\to a} f(x)$ and $f(a)$ had to be the same every time. Only some of the time.
A: Although this is a very very non rigours approach some people may kill me, but 
I will try to make this approach like this, you said why the function doesn't approaches same value.
$f(x)=\frac{x^2-4}{x-2}$
We are evaluating this at $x=2$,   
$f(x)=\frac{2^2-2^2}{2-2}$ 
It can be rewritten as, 
$f(x)=\frac{(2+2)2(1-1)}{2(1-1)}$
$f(x)=4 \frac{0}{0}$
So, the $0's$ cancels each other and yay same result, now why is this wrong?
Because we can write $\frac{0}{0}=\frac{100-100}{10-10}=10 $ or any other value think of! , that's why the above solution is not correct!
What limit does?
Well essentially it redefines $0$ a variable $h$ which $\to $ $0$ or any variable which introduces the undefined things. 
Now, we do not say this variable is 0, but it goes to 0 and never actually touches it. In the enlightenment of this new definition we will re-evaluate our function using this new definition. In your function it is $\lim_{x \to 2}(x-2)=\lim_{h \to 0} h$
$L=\lim_{x \to 2} \frac{(x+2)(x-2)}{(x-2)}$
$L= \lim_{x \to 2} \frac{(x+2)}{1} $
$L=4$
And there we didn't have to go through any undefined nasty 0 and problem solved. You can say limit saved the day!
