Continuity and types of discontinuity

Problem 1.190 · hard

1. A rational function is continuous wherever its denominator is not zero, so only the zeros of the denominator can be discontinuities.
   Reviewed
2. \[ x^{2} - 2 x - 15 = \left(x - 5\right) \left(x + 3\right) \]
   Factor the denominator.✓ Proved
3. \[ \frac{2 x^{2} - 18 x + 40}{x^{2} - 2 x - 15} = \frac{2 x - 8}{x + 3} \]
   The factor (x − 5) cancels.✓ Proved
4. \[ \lim_{x \to 5^+}\left(\frac{2 x^{2} - 18 x + 40}{x^{2} - 2 x - 15}\right) = \frac{1}{4} \]
   At x = 5 the limit exists, but f(5) is undefined: a removable discontinuity.✓ Proved
5. \[ \left. 2 x - 8 \right|_{\substack{ x=-3 }} = -14 \]
   At x = -3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved