Evaluate each of the following limits, then identify any vertical asymptotes of the function *f*(*x*)=1(*x*+1)2.

lim*x*→−1−1(*x*+1)2lim*x*→−1+1(*x*+1)2lim*x*→−11(*x*+1)2

If any of the limits do not exist, enter ∅ as your answer.




Let's evaluate each of the limits and analyze the function \( f(x) = \frac{1}{(x + 1)^2} \).

### 1. \( \lim_{x \to -1^-} \frac{1}{(x + 1)^2} \)

- As \( x \to -1^- \), \( x + 1 \to 0^- \), meaning that \( (x + 1) \) is approaching 0 from the negative side.
- Since \( (x + 1)^2 \) is squared, it will always be positive, regardless of whether \( x + 1 \) is positive or negative.
- Thus, \( (x + 1)^2 \to 0^+ \) as \( x \to -1^- \), and therefore \( \frac{1}{(x + 1)^2} \to +\infty \).

**So, \( \lim_{x \to -1^-} \frac{1}{(x + 1)^2} = +\infty \).**

### 2. \( \lim_{x \to -1^+} \frac{1}{(x + 1)^2} \)

- As \( x \to -1^+ \), \( x + 1 \to 0^+ \), meaning that \( (x + 1) \) is approaching 0 from the positive side.
- Again, \( (x + 1)^2 \to 0^+ \), so \( \frac{1}{(x + 1)^2} \to +\infty \).

**So, \( \lim_{x \to -1^+} \frac{1}{(x + 1)^2} = +\infty \).**

### 3. \( \lim_{x \to -1} \frac{1}{(x + 1)^2} \)

- Since the function tends to \( +\infty \) from both sides (left and right), the limit as \( x \to -1 \) does not exist in a finite form, but instead diverges to infinity.
- Therefore, we conclude that this limit does not exist in a finite sense.

**So, \( \lim_{x \to -1} \frac{1}{(x + 1)^2} = +\infty \) (or it does not exist in the usual sense).**

### Vertical Asymptote

The function \( f(x) = \frac{1}{(x + 1)^2} \) has a vertical asymptote at \( x = -1 \) because the denominator \( (x + 1)^2 \) goes to 0 as \( x \to -1 \), and the function approaches infinity at this point.

### Final Answers:

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

### Vertical Asymptote: \( x = -1 \)

Would you like further details or clarifications on any part? Here are some related questions:

1. What happens to the function \( f(x) = \frac{1}{(x + 1)^2} \) as \( x \) approaches other values?
2. How can you determine the behavior of functions near vertical asymptotes?
3. What other types of discontinuities exist, apart from vertical asymptotes?
4. How does the square of the denominator affect the behavior of the function near a vertical asymptote?
5. How do horizontal asymptotes differ from vertical asymptotes?

**Tip:** When analyzing functions near vertical asymptotes, always check the direction from both sides (left and right limits) to determine the function's behavior.

Evaluate each of the following limits, then identify any vertical asymptotes of the function f(x) = 1/(x+1)^2.

lim(x→−1−) 1/(x+1)^2
lim(x→−1+) 1/(x+1)^2
lim(x→−1) 1/(x+1)^2
If any of the limits do not exist, enter ∅ as your answer.

Learn how to evaluate limits approaching -1 and identify vertical asymptotes for the function f(x) = 1/(x+1)^2. This problem focuses on understanding limits from both sides of a point and recognizing vertical asymptotes in rational functions. Suitable for students in Grades 11-12.

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