The limit problem you've given is:

$\lim_{x \to -3^+} \frac{x^2 + 4}{x^2 - 9}$

Let's solve this step by step.

Step 1: Factor the denominator

The denominator $x^2 - 9$ can be factored using the difference of squares:

$x^2 - 9 = (x - 3)(x + 3)$

Now the expression becomes:

$\lim_{x \to -3^+} \frac{x^2 + 4}{(x - 3)(x + 3)}$

Step 2: Check the limit as $x \to -3^+$

When $x \to -3^+$, the behavior of each part of the fraction must be considered:

• The numerator $x^2 + 4$ approaches $9 + 4 = 13$.
• The factor $x - 3$ approaches $-6$ because $x$ is slightly greater than $-3$ from the right.
• The factor $x + 3$ approaches $0$, but since $x \to -3^+$, $x + 3$ is slightly positive.

Step 3: Analyze the behavior of the limit

Since the numerator approaches 13 (a finite number) and the denominator approaches 0, we analyze the sign of the denominator. The term $x + 3$ is positive when approaching from the right of $-3$, and $x - 3$ is negative. This means the denominator is approaching 0 from the negative side.

Thus, the limit can be interpreted as:

$\lim_{x \to -3^+} \frac{13}{\text{small negative number}} = -\infty$

Final Answer:

$\lim_{x \to -3^+} \frac{x^2 + 4}{x^2 - 9} = -\infty$

Would you like more details on any step, or do you have any questions?

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Here are 5 related questions for further practice:

1. $\lim_{x \to 3^-} \frac{x^2 + 4}{x^2 - 9}$
2. $\lim_{x \to -3} \frac{x^2 - 4}{x^2 - 9}$
3. $\lim_{x \to 0} \frac{x^2 + 1}{x^2 - 1}$
4. $\lim_{x \to -3^+} \frac{x^2 + 9}{x^2 - 9}$
5. $\lim_{x \to 2^+} \frac{x^2 - 4}{x^2 - 1}$

Tip: Always check the sign of the denominator when it approaches 0 to determine whether the limit tends toward $+\infty$ or $-\infty$.