Classifying the discontinuity of a rational function

First check the three conditions for continuity at $x=5$. Since substituting $x=5$ gives division by zero, $u(5)$ is undefined, so the first condition already fails. This alone tells us the function is discontinuous at $x=5$.

Next consider the behavior of $u(x)$ near $x=5$: as $x \to 5^-$, the denominator $x-5$ is a small negative number, so $u(x) \to -\infty$; as $x \to 5^+$, the denominator is a small positive number, so $u(x) \to +\infty$. Since the one-sided limits are infinite rather than finite (and unequal but not both finite), this corresponds to a vertical asymptote at $x=5$, so the discontinuity is classified as infinite, not removable or jump.

The first condition fails: u of five is undefined because plugging in five causes division by zero. Since the function value does not even exist at that point, we never get to check whether the limit exists or matches it. This is why the point is excluded from the domain entirely.