What is the limit of the following? $$ \lim _{x \rightarrow-3^{-}} \frac{x^{2}+14 x+45}{x+3} $$

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To evaluate the limit:

$$
\lim_{x \rightarrow -3^{-}} \frac{x^{2} + 14x + 45}{x + 3}
$$

**Step 1: Factor the Numerator**

Factor the quadratic expression in the numerator:

$$
x^2 + 14x + 45 = (x + 9)(x + 5)
$$

So, the expression becomes:

$$
\frac{(x + 9)(x + 5)}{x + 3}
$$

**Step 2: Analyze the Behavior as $$ x $$ Approaches $$-3$$ from the Left**

As $$ x $$ approaches $$-3$$ from the left ($$ x \rightarrow -3^- $$):

- **Numerator Evaluation:**
  $$
  (x + 9)(x + 5) \rightarrow (-3 + 9)(-3 + 5) = 6 \times 2 = 12
  $$
  
- **Denominator Evaluation:**
  $$
  x + 3 \rightarrow -3 + 3 = 0
  $$
  However, since $$ x $$ approaches $$-3$$ from the left, $$ x + 3 $$ approaches $$ 0 $$ **from the negative side** (i.e., $$ x + 3 $$ is slightly less than 0).

**Step 3: Determine the Limit**

Near $$ x = -3 $$, the expression behaves like:

$$
\frac{12}{x + 3}
$$

Since $$ x + 3 $$ approaches $$ 0 $$ from the negative side, the denominator is a very small negative number. Thus:

$$
\frac{12}{\text{a very small negative number}} \rightarrow -\infty
$$

**Conclusion:**

$$
\lim_{x \rightarrow -3^{-}} \frac{x^{2} + 14x + 45}{x + 3} = -\infty
$$

**Answer:** The limit is minus infinity.
The limit is minus infinity.

What is the limit of the following?

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