<s>[INST] 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 $.
 [/INST] It simply means that $f(x)$ approaches $4$ from both sides as $x$ approaches $2$. The function does not need to be defined at that point, although it could. The function $f(x) = \frac{x^2-4}{x-2}$ is exactly the "same" as the function $f(x) = x+2$ at all points except $x = 2$, where there is a removable discontinuity, or a gap, in the former function. Therefore, whatever value $x+2$ takes at $x=2$ will be equivalent to the value $\frac{x^2-4}{x-2}$ approaches as $x \to 2$. The fact that the function is undefined at that point doesn't change this.
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