Classify the discontinuity of w(x) at x = 5

First check condition 1: is $w(5)$ defined? Since the denominator $x-5$ becomes $0$ when $x=5$, the expression $\frac{1}{0}$ is undefined, so $w(5)$ does not exist. This already tells us $w$ is discontinuous at $x=5$.

Next, examine the behavior of $w(x)$ near $x=5$: as $x \to 5^-$, $x-5$ is a small negative number, so $w(x) \to -\infty$; as $x\to 5^+$, $x-5$ is a small positive number, so $w(x)\to +\infty$. Since at least one one-sided limit is infinite (rather than approaching a finite value), $\lim_{x\to5} w(x)$ does not exist as a real number, and this behavior corresponds to a vertical asymptote at $x=5$. Therefore the discontinuity at $x=5$ is classified as an infinite discontinuity (condition 1, $f(a)$ defined, fails, and the limit fails to exist for this reason).

The first condition fails: w of five is not defined, since plugging in five gives division by zero. Because this condition already fails, the function cannot be continuous at that point regardless of the other two conditions. This matches the earlier finding that w is undefined at x equals five before any limit behavior is even considered.