Question
for each function, determine the long run behavior.
x^3+1 / x^2+2 has
no horizontal asympotote
a horizontal asymptote at y=0
a horizontal asymptote at y=1
x^2+1/ x^3+2 has
no horizontal asymptote
a horizontal asymptote at y=0
a horizontal asymptote at y=1
x^2+1 / x^2+2 has
no horizontal asymptote
a horizontal asymptote at y=0
a horizontal asymptote at y=1
x^3+1 / x^2+2 has
no horizontal asympotote
a horizontal asymptote at y=0
a horizontal asymptote at y=1
x^2+1/ x^3+2 has
no horizontal asymptote
a horizontal asymptote at y=0
a horizontal asymptote at y=1
x^2+1 / x^2+2 has
no horizontal asymptote
a horizontal asymptote at y=0
a horizontal asymptote at y=1
Answers
GPT-4o mini
To determine the long-run behavior (horizontal asymptotes) of the functions you provided, we need to analyze the limits of the given functions as \( x \) approaches infinity (\( x \to \infty \)) and negative infinity (\( x \to -\infty \)).
### 1. Function: \(\frac{x^3 + 1}{x^2 + 2}\)
To find the horizontal asymptote, we evaluate the limit:
\[
\lim_{x \to \infty} \frac{x^3 + 1}{x^2 + 2}
\]
- The highest power in the numerator is \( x^3 \) and in the denominator is \( x^2 \). Therefore, divide every term by \( x^2 \):
\[
= \lim_{x \to \infty} \frac{x^3/x^2 + 1/x^2}{x^2/x^2 + 2/x^2} = \lim_{x \to \infty} \frac{x + 0}{1 + 0} = \infty
\]
**Conclusion:** There is **no horizontal asymptote**.
---
### 2. Function: \(\frac{x^2 + 1}{x^3 + 2}\)
Again, we evaluate the limit:
\[
\lim_{x \to \infty} \frac{x^2 + 1}{x^3 + 2}
\]
- The highest power in the numerator is \( x^2 \) and in the denominator is \( x^3 \). Therefore, divide every term by \( x^3 \):
\[
= \lim_{x \to \infty} \frac{x^2/x^3 + 1/x^3}{x^3/x^3 + 2/x^3} = \lim_{x \to \infty} \frac{1/x + 0}{1 + 2/x^3} = \lim_{x \to \infty} \frac{0}{1} = 0
\]
**Conclusion:** There is a horizontal asymptote at \( y = 0 \).
---
### 3. Function: \(\frac{x^2 + 1}{x^2 + 2}\)
Now, evaluate this limit as well:
\[
\lim_{x \to \infty} \frac{x^2 + 1}{x^2 + 2}
\]
- The highest power in both numerator and denominator is \( x^2 \). So we divide every term by \( x^2 \):
\[
= \lim_{x \to \infty} \frac{x^2/x^2 + 1/x^2}{x^2/x^2 + 2/x^2} = \lim_{x \to \infty} \frac{1 + 0}{1 + 0} = 1
\]
**Conclusion:** There is a horizontal asymptote at \( y = 1 \).
---
### Summary of Results:
1. **\(\frac{x^3 + 1}{x^2 + 2}\)**: No horizontal asymptote.
2. **\(\frac{x^2 + 1}{x^3 + 2}\)**: Horizontal asymptote at \( y = 0 \).
3. **\(\frac{x^2 + 1}{x^2 + 2}\)**: Horizontal asymptote at \( y = 1 \).
### 1. Function: \(\frac{x^3 + 1}{x^2 + 2}\)
To find the horizontal asymptote, we evaluate the limit:
\[
\lim_{x \to \infty} \frac{x^3 + 1}{x^2 + 2}
\]
- The highest power in the numerator is \( x^3 \) and in the denominator is \( x^2 \). Therefore, divide every term by \( x^2 \):
\[
= \lim_{x \to \infty} \frac{x^3/x^2 + 1/x^2}{x^2/x^2 + 2/x^2} = \lim_{x \to \infty} \frac{x + 0}{1 + 0} = \infty
\]
**Conclusion:** There is **no horizontal asymptote**.
---
### 2. Function: \(\frac{x^2 + 1}{x^3 + 2}\)
Again, we evaluate the limit:
\[
\lim_{x \to \infty} \frac{x^2 + 1}{x^3 + 2}
\]
- The highest power in the numerator is \( x^2 \) and in the denominator is \( x^3 \). Therefore, divide every term by \( x^3 \):
\[
= \lim_{x \to \infty} \frac{x^2/x^3 + 1/x^3}{x^3/x^3 + 2/x^3} = \lim_{x \to \infty} \frac{1/x + 0}{1 + 2/x^3} = \lim_{x \to \infty} \frac{0}{1} = 0
\]
**Conclusion:** There is a horizontal asymptote at \( y = 0 \).
---
### 3. Function: \(\frac{x^2 + 1}{x^2 + 2}\)
Now, evaluate this limit as well:
\[
\lim_{x \to \infty} \frac{x^2 + 1}{x^2 + 2}
\]
- The highest power in both numerator and denominator is \( x^2 \). So we divide every term by \( x^2 \):
\[
= \lim_{x \to \infty} \frac{x^2/x^2 + 1/x^2}{x^2/x^2 + 2/x^2} = \lim_{x \to \infty} \frac{1 + 0}{1 + 0} = 1
\]
**Conclusion:** There is a horizontal asymptote at \( y = 1 \).
---
### Summary of Results:
1. **\(\frac{x^3 + 1}{x^2 + 2}\)**: No horizontal asymptote.
2. **\(\frac{x^2 + 1}{x^3 + 2}\)**: Horizontal asymptote at \( y = 0 \).
3. **\(\frac{x^2 + 1}{x^2 + 2}\)**: Horizontal asymptote at \( y = 1 \).