To evaluate the left-hand limit:
$\lim_{x \to -3^-} \frac{f(x) + 2}{x + 3}$ we need to analyze the behavior of $f(x)$ as $x$ approaches $-3$ from the left.

Step 1: Find $\lim_{x \to -3^-} f(x)$

From the graph, observe the left-hand limit of $f(x)$ at $x = -3$. The function approaches $-1$ from the left.

Thus,
$\lim_{x \to -3^-} f(x) = -1.$

Step 2: Compute the limit of the fraction

Substituting $f(x)$ into the given expression:

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

$= \lim_{x \to -3^-} \frac{1}{x + 3}.$

Step 3: Evaluate the fraction

As $x \to -3^-$, the denominator $x + 3$ approaches 0 from the negative side (since $x$ is slightly less than -3). This makes $\frac{1}{x + 3}$ approach negative infinity.

Conclusion:

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

Would you like a more detailed explanation or additional help?

Related Questions:

1. What is $\lim_{x \to -3^+} \frac{f(x) + 2}{x + 3}$?
2. How do we determine the limit from the right at $x = -3$?
3. What are the discontinuities in $f(x)$ based on the graph?
4. How can we interpret the behavior of $\frac{f(x) + 2}{x + 3}$ at other points?
5. What happens if we evaluate $\lim_{x \to -3} \frac{f(x) + 2}{x + 3}$ as a two-sided limit?

Tip:

When evaluating limits involving fractions, always check if the denominator approaches 0 to determine if the function goes to infinity or negative infinity.