To estimate the value of the limit $\lim_{x \to -3} \frac{x^2 - 3x}{x^2 - 9}$, we can evaluate the function at points increasingly close to $x = -3$ from both the left and right sides.

The function to evaluate is:

$f(x) = \frac{x^2 - 3x}{x^2 - 9}$

We need to evaluate this function at the given values of $x$: $-2.5$, $-2.9$, $-2.95$, $-2.99$, $-2.999$, $-2.9999$, $-3.5$, $-3.1$, $-3.05$, $-3.01$, $-3.001$, and $-3.0001$.

Let me compute the values of $f(x)$ at these points.The values of the function $f(x) = \frac{x^2 - 3x}{x^2 - 9}$ at the given points are as follows:

• $f(-2.5) = -5.000000$
• $f(-2.9) = -29.000000$
• $f(-2.95) = -59.000000$
• $f(-2.99) = -299.000000$
• $f(-2.999) = -2999.000000$
• $f(-2.9999) = -29999.000000$
• $f(-3.5) = 7.000000$
• $f(-3.1) = 31.000000$
• $f(-3.05) = 61.000000$
• $f(-3.01) = 301.000000$
• $f(-3.001) = 3001.000000$
• $f(-3.0001) = 30001.000000$

As $x$ approaches $-3$ from the left, the function values become increasingly large negative numbers. As $x$ approaches $-3$ from the right, the function values become increasingly large positive numbers. This suggests that the function has a vertical asymptote at $x = -3$, and therefore the limit does not exist.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the general approach to finding the limit of a rational function?
2. How do vertical asymptotes affect the existence of a limit?
3. Can you graph the function to visualize the behavior near $x = -3$?
4. What happens to the limit if we approach $x = 3$ instead of $x = -3$?
5. How can we algebraically determine the presence of a vertical asymptote in a function?

Tip: When evaluating limits, always consider the behavior of the function from both the left and right sides of the point in question to identify potential asymptotes or discontinuities.