Let's analyze the limits of the given function $f(x) = -\frac{3(x+2)}{x^2 + 4x + 4}$ at $x = -2$.

The function has a vertical asymptote at $x = -2$. This is because the denominator $x^2 + 4x + 4$ becomes 0 when $x = -2$. To find each limit, we'll evaluate the behavior of the function from both the left-hand side ($x \to -2^-$) and the right-hand side ($x \to -2^+$).

1. Limit from the left $\lim_{x \to -2^-} \left( -\frac{3(x+2)}{x^2 + 4x + 4} \right)$

The denominator is $x^2 + 4x + 4 = (x + 2)^2$. At $x = -2$, this denominator is 0 because $(x + 2)^2 = 0$. Since the denominator is squared, it will always be positive as we approach $x = -2$ from either side, but the numerator depends on the behavior of $x + 2$.

• As $x \to -2^-$, the term $x + 2$ becomes a small negative number (just slightly less than 0).
• Therefore, $-3(x+2)$ becomes a small positive number (since the negative of a negative is positive).

Hence, as $x \to -2^-$, the function behaves like:

$\lim_{x \to -2^-} -\frac{3(x+2)}{(x+2)^2} \approx \frac{\text{positive small number}}{\text{positive small number}} \to +\infty$

Thus:

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

2. Limit from the right $\lim_{x \to -2^+} \left( -\frac{3(x+2)}{x^2 + 4x + 4} \right)$

Now, let's approach $x = -2$ from the right-hand side ($x \to -2^+$).

• As $x \to -2^+$, the term $x + 2$ becomes a small positive number (just slightly greater than 0).
• Therefore, $-3(x+2)$ becomes a small negative number.

Thus, as $x \to -2^+$, the function behaves like:

$\lim_{x \to -2^+} -\frac{3(x+2)}{(x+2)^2} \approx \frac{\text{negative small number}}{\text{positive small number}} \to -\infty$

Therefore:

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

3. Two-sided limit $\lim_{x \to -2} \left( -\frac{3(x+2)}{x^2 + 4x + 4} \right)$

Since the left-hand limit and right-hand limit as $x \to -2$ do not match (one approaches $+\infty$ and the other approaches $-\infty$), the two-sided limit does not exist.

Thus:

$\lim_{x \to -2} \left( -\frac{3(x+2)}{x^2 + 4x + 4} \right) = \text{does not exist}$

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Summary:

1. $\lim_{x \to -2^-} \left( -\frac{3(x+2)}{x^2 + 4x + 4} \right) = +\infty$
2. $\lim_{x \to -2^+} \left( -\frac{3(x+2)}{x^2 + 4x + 4} \right) = -\infty$
3. $\lim_{x \to -2} \left( -\frac{3(x+2)}{x^2 + 4x + 4} \right) = \text{does not exist}$

Would you like further clarification or details on any of these steps?

Related Questions:

1. How do vertical asymptotes affect the graph of a rational function?
2. Can a function have more than one vertical asymptote?
3. What is the difference between vertical and horizontal asymptotes?
4. How do you find horizontal asymptotes of a rational function?
5. What is the behavior of a function near a hole versus an asymptote?

Tip:

When dealing with rational functions, always factor the denominator and numerator completely to help locate vertical asymptotes and simplify limits.