Why does $ \lim_{x \to 2} \frac{x^2-4}{x-2} =4 $ if x cannot be 2? I know that $ \lim_{x \to 2} \frac{x^2-4}{x-2} $ is evaluated as follows :-
$$ \lim_{x \to 2} \frac{x^2-4}{x-2} \\ = \lim_{x \to 2} \frac{(x+2)(x-2)}{x-2} \\ = \lim_{x \to 2} x+2 \\ = 2+2 \\ = 4 $$
By looking at the function $ \frac{x^2 - 4}{x - 2} $, I can see that 2 is not in its domain. Therefore, I am not able to understand how $ \lim_{x \to 2} x+2 = 2+2 $.

Limit doesn't have to do with the existence of function at that point. In the context of the given problem, 
$$f(x) = \frac{x^2-4}{x-2}= \frac{(x-2)(x+2)}{x-2}$$
We study the behavior of $f$ as $x$ approaches a certain point. For example let's study the RHL by considering the value of $f$ at some points: 
$$f(2.01) = 4.01$$
$$f(2.001) = 4.001$$
$$f(2.0001) = 4.0001$$
You see how when we move $x$ closer to $2$, $f(x)$ gets closer to $4$. 
Similarly for the LHL, 
$$f(1.99) = 3.99$$
$$f(1.999) = 3.999$$
$$f(1.9999) = 3.9999$$
The same thing happens when we approach $x$ closer to $2$, $f(x)$ gets closer to $4$. 
Now in short hand we can do it like this
$$\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}$$
Now since $ x \neq 2$, $x - 2 \neq 0$. Therefore we can cancel the factor in the numerator and denominator above. Hence we get
$$\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}$$
$$=\lim_{x \to 2} x+2$$
$$=4$$
