Continuity and types of discontinuity

Problem 1.189 · 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. \[ x^{2} - 7 x + 10 = \left(x - 5\right) \left(x - 2\right) \]
   Factor the denominator.✓ Proved
3. \[ \frac{x^{2} - 25}{x^{2} - 7 x + 10} = \frac{x + 5}{x - 2} \]
   The factor (x − 5) cancels.✓ Proved
4. \[ \lim_{x \to 5^+}\left(\frac{x^{2} - 25}{x^{2} - 7 x + 10}\right) = \frac{10}{3} \]
   At x = 5 the limit exists, but f(5) is undefined: a removable discontinuity.✓ Proved
5. \[ \left. x + 5 \right|_{\substack{ x=2 }} = 7 \]
   At x = 2 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved