To solve the limit of $\frac{x}{\sqrt{x^2 - 9}}$ as $x \to -3^{-}$, we proceed step-by-step:

Step 1: Examine the behavior of the function as $x \to -3$ from the left.

We have the expression: $f(x) = \frac{x}{\sqrt{x^2 - 9}}.$

Step 2: Analyze the denominator.

In the denominator, $\sqrt{x^2 - 9}$, the expression $x^2 - 9$ becomes 0 when $x = \pm 3$. So, near $x = -3$, we have: $x^2 - 9 = (-3)^2 - 9 = 9 - 9 = 0.$ As $x \to -3^{-}$, the expression $x^2 - 9$ is positive but very small, so $\sqrt{x^2 - 9}$ approaches 0 from the positive side.

Step 3: Analyze the numerator.

The numerator is simply $x$, and as $x \to -3^{-}$, $x$ approaches -3.

Step 4: Evaluate the limit.

Now, we need to consider the behavior of the whole expression: $\frac{x}{\sqrt{x^2 - 9}}.$ As $x \to -3^{-}$:

• The numerator approaches $-3$.
• The denominator approaches 0 from the positive side (since $x^2 - 9$ is positive and small near $x = -3$).

Thus, the fraction $\frac{-3}{\text{small positive number}}$ becomes very large and negative. Therefore, the limit is: $\lim_{x \to -3^{-}} \frac{x}{\sqrt{x^2 - 9}} = -\infty.$

Final Answer:

$\lim_{x \to -3^{-}} \frac{x}{\sqrt{x^2 - 9}} = -\infty.$

Would you like further details or have any other questions?

Here are 5 related questions for further exploration:

1. What is the limit of $\frac{x}{\sqrt{x^2 - 9}}$ as $x \to 3^+$?
2. How do you evaluate limits involving radicals and absolute values?
3. Can we simplify expressions with radicals to find limits more easily?
4. What happens to the function $\frac{x}{\sqrt{x^2 - 9}}$ at $x = 0$?
5. What are one-sided limits, and why are they important?

Tip: When evaluating limits with square roots, always check the sign of the expression under the square root to ensure you are considering valid real numbers.