For a quotient $\dfrac{N(x)}{D(x)}$, where $N$ is the numerator and $D$ the denominator, the first move at $x \to a$ is always direct substitution. If $D(a) \ne 0$, the quotient is a well-behaved function near $a$ and $\dfrac{N(a)}{D(a)}$ IS the limit: done. If $N(a) \ne 0$ and $D(a) = 0$, the values blow up near $a$: a vertical asymptote, no finite limit. And if both are zero, you have $\dfrac{0}{0}$, the indeterminate form. It is not zero, not one, not undefined, not a verdict of any kind: it is the form telling you that $N$ and $D$ share a factor $(x - a)$, which is the only way both can vanish at the same place.

So factor both, cancel the common $(x - a)$, and substitute again: $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2} = \lim_{x \to 2} \dfrac{(x-2)(x+2)}{x-2} = \lim_{x \to 2}(x + 2) = 4$, read aloud: the limit as x approaches two of x squared minus four over x minus two equals four. Cancelling is legal precisely because the limit never evaluates AT $x = 2$: everywhere else the two quotients are identical, and everywhere else is all a limit looks at.