Evaluating an infinite limit at x = -2

Since $x \neq 7$ near $x=-2$, we can cancel the common factor: $f(x) = \dfrac{x-7}{(x-7)(x+2)} = \dfrac{1}{x+2}$ for $x \neq 7$. Note that the cancellation is valid for all $x$ in a neighborhood of $x = -2$ (where $x \neq 7$ is automatically satisfied), so we can use the simplified form to analyze the limit.

As $x \to -2^+$, $x+2 \to 0^+$, so $\dfrac{1}{x+2} \to +\infty$. As $x \to -2^-$, $x+2 \to 0^-$, so $\dfrac{1}{x+2} \to -\infty$. Since the one-sided limits disagree, the two-sided limit does not exist. A vertical asymptote occurs at $x = -2$ because $f$ is undefined there and at least one one-sided limit is infinite; both conditions are satisfied here.