A function is said to have a vertical asymptote wherever the limit on the left or right (or both) is either positive or negative infinity. 
For example, the function f(x)=−3(x+2)x2+4x+4 has a vertical asymptote at x=−2.
Find each of the following limits.
limx→−2−−3(x+2)x2+4x+4= help (limits)

limx→−2+−3(x+2)x2+4x+4= help (limits)

limx→−2−3(x+2)x2+4x+4=  help (limits)

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:

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

---

### 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.

A function is said to have a vertical asymptote wherever the limit on the left or right (or both) is either positive or negative infinity. For example, the function f(x)=−3(x+2)/(x^2+4x+4) has a vertical asymptote at x=−2. Find each of the following limits: lim x→−2− −3(x+2)/(x^2+4x+4), lim x→−2+ −3(x+2)/(x^2+4x+4), lim x→−2 −3(x+2)/(x^2+4x+4).

Explore the detailed solution to finding limits of the rational function f(x) = -3(x+2)/(x^2+4x+4) at x = -2, where the function has a vertical asymptote. Learn about the behavior of limits as x approaches from both the left and the right. Suitable for Grades 10-12 students studying calculus concepts like limits and asymptotes.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation