Continuity and types of discontinuity

Problem 1.193 · 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. \[ x^{2} + 3 x - 4 = \left(x - 1\right) \left(x + 4\right) \]
   Factor the denominator.✓ Proved
3. \[ \frac{- 2 x^{2} - 14 x - 24}{x^{2} + 3 x - 4} = \frac{- 2 x - 6}{x - 1} \]
   The factor (x + 4) cancels.✓ Proved
4. \[ \lim_{x \to -4^+}\left(\frac{- 2 x^{2} - 14 x - 24}{x^{2} + 3 x - 4}\right) = - \frac{2}{5} \]
   At x = -4 the limit exists, but f(-4) is undefined: a removable discontinuity.✓ Proved
5. \[ \left. - 2 x - 6 \right|_{\substack{ x=1 }} = -8 \]
   At x = 1 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved