Continuity and types of discontinuity

Problem 1.198 · 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. \[ x^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right) \]
   Factor the denominator.✓ Proved
3. \[ \frac{2 x^{2} + 4 x - 16}{x^{2} - x - 20} = \frac{2 x - 4}{x - 5} \]
   The factor (x + 4) cancels.✓ Proved
4. \[ \lim_{x \to -4^+}\left(\frac{2 x^{2} + 4 x - 16}{x^{2} - x - 20}\right) = \frac{4}{3} \]
   At x = -4 the limit exists, but f(-4) is undefined: a removable discontinuity.✓ Proved
5. \[ \left. 2 x - 4 \right|_{\substack{ x=5 }} = 6 \]
   At x = 5 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved