One-sided limits and existence of the two-sided limit at a removable discontinuity

For $x \neq 2$, factor the numerator: $\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2$. This simplification is valid for all $x$ near $2$ except exactly at $x=2$, which is fine since limits only care about nearby values.

As $x \to 2^-$, $x+2 \to 4$, so $\lim_{x\to 2^-} f(x) = 4$. Similarly, as $x \to 2^+$, $x+2 \to 4$, so $\lim_{x\to 2^+} f(x) = 4$. Since both one-sided limits equal $4$, the two-sided limit exists and equals $4$ — note this differs from $f(2)=5$, illustrating that the limit does not depend on the function's actual value at $x=2$.