Question: For the rational functions below, answer each of the following questions.
$f(x) = \frac{x - 1}{x^2 + 4x - 5}$
(Note: A discontinuity is called removable if there is a hole in the graph, and it is called infinite if there is a vertical asymptote.)
Provide your answer below:
(a) $\lim_{x \to -5^-} f(x) =$
(b) $\lim_{x \to -5^+} f(x) =$
(c) $f$ has a(n) discontinuity at $x = -5$. (removable/infinite)
1. **State the problem:** We need to find the left-hand limit, right-hand limit of $f(x) = \frac{x - 1}{x^2 + 4x - 5}$ as $x$ approaches $-5$, and classify the discontinuity at $x = -5$.
2. **Factor the denominator:**
$$x^2 + 4x - 5 = (x + 5)(x - 1)$$
3. **Rewrite the function:**
$$f(x) = \frac{x - 1}{(x + 5)(x - 1)}$$
4. **Simplify the function where possible:**
For $x \neq 1$, we can cancel $x - 1$:
$$f(x) = \frac{\cancel{x - 1}}{(x + 5)\cancel{(x - 1)}} = \frac{1}{x + 5}$$
5. **Evaluate the limits as $x$ approaches $-5$:**
- Left-hand limit ($x \to -5^-$):
$$\lim_{x \to -5^-} f(x) = \lim_{x \to -5^-} \frac{1}{x + 5} = \frac{1}{-5 + 5^-} = \frac{1}{0^-} = -\infty$$
- Right-hand limit ($x \to -5^+$):
$$\lim_{x \to -5^+} f(x) = \lim_{x \to -5^+} \frac{1}{x + 5} = \frac{1}{0^+} = +\infty$$
6. **Classify the discontinuity at $x = -5$:**
Since the limits approach infinity (one side $-\infty$, the other $+\infty$), there is a vertical asymptote at $x = -5$. This is an infinite discontinuity.
**Final answers:**
(a) $\lim_{x \to -5^-} f(x) = -\infty$
(b) $\lim_{x \to -5^+} f(x) = +\infty$
(c) $f$ has an infinite discontinuity at $x = -5$.