The given function is:

$f(x) = \frac{x^2 - 8x + 15}{x^2 - 3x}$

To analyze the limits, let's simplify the function first and then evaluate the required limits at the points specified.

Step 1: Simplify the Function

1. Factor the numerator and denominator:

   • The numerator $x^2 - 8x + 15$ can be factored as $(x - 3)(x - 5)$.
   • The denominator $x^2 - 3x$ can be factored as $x(x - 3)$.

   So, $f(x) = \frac{(x - 3)(x - 5)}{x(x - 3)}$

2. Cancel common terms: For $x \neq 3$, we can cancel $(x - 3)$ from the numerator and the denominator, resulting in: $f(x) = \frac{x - 5}{x}, \quad x \neq 3$

Step 2: Evaluate the Limits

(a) $\lim_{x \to 0^-} f(x)$

Using the simplified form $f(x) = \frac{x - 5}{x}$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x - 5}{x} = \lim_{x \to 0^-} \left(1 - \frac{5}{x}\right)$ As $x \to 0^-$, $\frac{5}{x} \to -\infty$, so $f(x) \to +\infty$. $\lim_{x \to 0^-} f(x) = +\infty$

(b) $\lim_{x \to 0^+} f(x)$

Similarly, $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(1 - \frac{5}{x}\right)$ As $x \to 0^+$, $\frac{5}{x} \to +\infty$, so $f(x) \to -\infty$. $\lim_{x \to 0^+} f(x) = -\infty$

(c) $\lim_{x \to 3^-} f(x)$

Since $x = 3$ is a point of discontinuity in the original form of $f(x)$, we use the simplified form: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{x - 5}{x} = \frac{3 - 5}{3} = \frac{-2}{3} = -\frac{2}{3}$

(d) $\lim_{x \to 3^+} f(x)$

Similarly, $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{x - 5}{x} = \frac{3 - 5}{3} = -\frac{2}{3}$

Summary of Limits

1. $\lim_{x \to 0^-} f(x) = +\infty$
2. $\lim_{x \to 0^+} f(x) = -\infty$
3. $\lim_{x \to 3^-} f(x) = -\frac{2}{3}$
4. $\lim_{x \to 3^+} f(x) = -\frac{2}{3}$

Would you like further clarification on any of these steps or have additional questions?

Here are five questions to deepen your understanding:

1. Why is $x = 3$ a point of discontinuity for $f(x)$?
2. What would happen if we calculated $\lim_{x \to \infty} f(x)$?
3. How does factor cancellation affect the behavior of a function near discontinuities?
4. Can limits at points of discontinuity have finite values?
5. What happens to $f(x)$ as $x$ approaches large negative values?

Tip: When simplifying functions, always check for points where terms might cancel, as these often indicate removable discontinuities or changes in behavior at certain points.