8. $\lim\limits_{x\to\infty} \frac{3x^2 + 20x}{4x^2 + 9}$
9. $\lim\limits_{x\to\infty} \frac{3x^2 + 20x}{2x^4 + 3x^3 - 29}$
10. $\lim\limits_{x\to\infty} \frac{4}{x + 5}$
11. $\lim\limits_{x\to\infty} \frac{7x - 9}{4x + 3}$
12. $\lim\limits_{x\to\infty} \frac{9x^2 - 2}{6 - 29x}$
13. $\lim\limits_{x\to-\infty} \frac{7x^2 - 9}{4x + 3}$
14. $\lim\limits_{x\to-\infty} \frac{5x - 9}{4x^3 + 2x + 7}$
15. $\lim\limits_{x\to-\infty} \frac{3x^3 - 10}{x + 4}$
16. $\lim\limits_{x\to-\infty} \frac{2x^5 + 3x^4 - 31x}{8x^4 - 31x^2 + 12}$
in exercises 17–24, find the horizontal asymptotes.

Question

Accepted Answer

Let's solve each limit problem one by one. We'll use the rule for limits at infinity for rational functions: if we have a rational function $\frac{f(x)}{g(x)}$ where $f(x)$ is a polynomial of degree $n$ and $g(x)$ is a polynomial of degree $m$, then:

If $n < m$, $\lim_{x \to \pm\infty} \frac{f(x)}{g(x)} = 0$
If $n = m$, $\lim_{x \to \pm\infty} \frac{f(x)}{g(x)} = \frac{\text{leading coefficient of } f(x)}{\text{leading coefficient of } g(x)}$
If $n > m$, $\lim_{x \to \pm\infty} \frac{f(x)}{g(x)} = \pm\infty$ (the sign depends on the leading coefficients and the sign of $x$ as $x \to \pm\infty$)

Problem 8: $\lim_{x \to \infty} \frac{3x^2 + 20x}{4x^2 + 9}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($3x^2 + 20x$): $2$
Degree of denominator ($4x^2 + 9$): $2$

Since the degrees are equal, we take the ratio of the leading coefficients.

Step 2: Take ratio of leading coefficients

Leading coefficient of numerator: $3$
Leading coefficient of denominator: $4$

So, $\lim_{x \to \infty} \frac{3x^2 + 20x}{4x^2 + 9} = \frac{3}{4}$

Problem 9: $\lim_{x \to \infty} \frac{3x^2 + 20x}{2x^4 + 3x^3 - 29}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($3x^2 + 20x$): $2$
Degree of denominator ($2x^4 + 3x^3 - 29$): $4$

Since $2 < 4$ (degree of numerator < degree of denominator), the limit is $0$.

So, $\lim_{x \to \infty} \frac{3x^2 + 20x}{2x^4 + 3x^3 - 29} = 0$

Problem 10: $\lim_{x \to \infty} \frac{4}{x + 5}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($4$): $0$ (constant term)
Degree of denominator ($x + 5$): $1$

Since $0 < 1$ (degree of numerator < degree of denominator), the limit is $0$.

So, $\lim_{x \to \infty} \frac{4}{x + 5} = 0$

Problem 11: $\lim_{x \to \infty} \frac{7x - 9}{4x + 3}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($7x - 9$): $1$
Degree of denominator ($4x + 3$): $1$

Since the degrees are equal, we take the ratio of the leading coefficients.

Step 2: Take ratio of leading coefficients

Leading coefficient of numerator: $7$
Leading coefficient of denominator: $4$

So, $\lim_{x \to \infty} \frac{7x - 9}{4x + 3} = \frac{7}{4}$

Problem 12: $\lim_{x \to \infty} \frac{9x^2 - 2}{6 - 29x}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($9x^2 - 2$): $2$
Degree of denominator ($6 - 29x$): $1$

Since $2 > 1$ (degree of numerator > degree of denominator), we analyze the sign. As $x \to \infty$, the leading term of the numerator is $9x^2$ (positive) and the leading term of the denominator is $-29x$ (negative for large $x$). So the limit will be $-\infty$ (because positive divided by negative is negative, and the magnitude goes to infinity).

So, $\lim_{x \to \infty} \frac{9x^2 - 2}{6 - 29x} = -\infty$

Problem 13: $\lim_{x \to -\infty} \frac{7x^2 - 9}{4x + 3}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($7x^2 - 9$): $2$
Degree of denominator ($4x + 3$): $1$

Since $2 > 1$ (degree of numerator > degree of denominator), we analyze the sign. As $x \to -\infty$:

Leading term of numerator: $7x^2$ (positive, since square of negative is positive)
Leading term of denominator: $4x$ (negative, since $x$ is negative)

So the ratio of leading terms is $\frac{7x^2}{4x} = \frac{7x}{4}$. As $x \to -\infty$, $\frac{7x}{4} \to -\infty$ (because $x$ is negative and magnitude goes to infinity).

Thus, $\lim_{x \to -\infty} \frac{7x^2 - 9}{4x + 3} = -\infty$

Problem 14: $\lim_{x \to -\infty} \frac{5x - 9}{4x^3 + 2x + 7}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($5x - 9$): $1$
Degree of denominator ($4x^3 + 2x + 7$): $3$

Since $1 < 3$ (degree of numerator < degree of denominator), the limit is $0$.

So, $\lim_{x \to -\infty} \frac{5x - 9}{4x^3 + 2x + 7} = 0$

Problem 15: $\lim_{x \to -\infty} \frac{3x^3 - 10}{x + 4}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($3x^3 - 10$): $3$
Degree of denominator ($x + 4$): $1$

Since $3 > 1$ (degree of numerator > degree of denominator), we analyze the sign. As $x \to -\infty$:

Leading term of numerator: $3x^3$ (negative, since cube of negative is negative)
Leading term of denominator: $x$ (negative, since $x$ is negative)

Ratio of leading terms: $\frac{3x^3}{x} = 3x^2$. As $x \to -\infty$, $x^2 \to \infty$, so $3x^2 \to \infty$ (positive, since square of negative is positive). Wait, wait, let's check again:

Wait, numerator leading term: $3x^3$ (as $x \to -\infty$, $x^3$ is negative, so $3x^3$ is negative)
Denominator leading term: $x$ (as $x \to -\infty$, $x$ is negative)
So $\frac{3x^3}{x} = 3x^2$. Wait, $x^2$ is positive, so $3x^2$ is positive. But wait, let's do it properly:

$\frac{3x^3 - 10}{x + 4} = \frac{x^3(3 - \frac{10}{x^3})}{x(1 + \frac{4}{x})} = \frac{x^2(3 - \frac{10}{x^3})}{1 + \frac{4}{x}}$

As $x \to -\infty$, $x^2 \to \infty$ (since square of negative is positive), and the other terms $\frac{10}{x^3} \to 0$ and $\frac{4}{x} \to 0$. So the limit is $\infty$? Wait, no, wait:

Wait, $x \to -\infty$, so $x^2$ is positive and goes to infinity. So $\frac{x^2(3 - 0)}{1 + 0} = 3x^2 \to \infty$ as $x \to -\infty$. Wait, but let's check the sign again. Wait, numerator: $3x^3$ when $x \to -\infty$ is negative (since $x^3$ is negative), denominator: $x$ when $x \to -\infty$ is negative. So negative divided by negative is positive. And the magnitude: $|3x^3| / |x| = 3x^2$, which goes to infinity. So the limit is $\infty$? Wait, no, wait:

Wait, $x \to -\infty$, so $x$ is negative, $x^3$ is negative, $x^2$ is positive. So:

$\frac{3x^3 - 10}{x + 4} \approx \frac{3x^3}{x} = 3x^2$ as $x \to -\infty$. Since $x^2$ is positive and goes to infinity, $3x^2$ goes to $\infty$. So the limit is $\infty$? Wait, but let's check with actual substitution. Let $x = -1000$:

Numerator: $3(-1000)^3 - 10 = 3(-10^9) - 10 = -3 \times 10^9 - 10$ (negative)
Denominator: $-1000 + 4 = -996$ (negative)
So negative divided by negative is positive. And the magnitude: $| -3 \times 10^9 | / | -996 | \approx 3 \times 10^6$, which is large. As $x$ becomes more negative, the magnitude increases. So the limit is $\infty$? Wait, but according to the degree rule: degree of numerator (3) > degree of denominator (1), so the limit is $\pm\infty$. The sign: leading coefficient of numerator is 3 (positive), leading coefficient of denominator is 1 (positive). But since $x \to -\infty$, the numerator's leading term is $3x^3$ (negative, because $x^3$ is negative), denominator's leading term is $x$ (negative). So positive (3) times negative (x^3) is negative, divided by negative (x) is positive. So the limit is $\infty$. Wait, but actually, when degree of numerator is greater than denominator, the limit is $\pm\infty$ depending on the sign. So here, since the leading terms give a positive result (negative/negative), the limit is $\infty$. Wait, but let's re-express:

$\lim_{x \to -\infty} \frac{3x^3 - 10}{x + 4} = \lim_{x \to -\infty} \frac{x^3(3 - \frac{10}{x^3})}{x(1 + \frac{4}{x})} = \lim_{x \to -\infty} x^2 \cdot \frac{3 - \frac{10}{x^3}}{1 + \frac{4}{x}}$

As $x \to -\infty$, $x^2 \to \infty$, and $\frac{3 - \frac{10}{x^3}}{1 + \frac{4}{x}} \to \frac{3 - 0}{1 + 0} = 3$. So the product is $\infty \times 3 = \infty$. So the limit is $\infty$? Wait, but that seems conflicting with my initial thought. Wait, no, $x^2$ as $x \to -\infty$ is positive and goes to infinity, so multiplying by 3 (positive) gives infinity. So the limit is $\infty$. Wait, but let's check with $x \to -\infty$, $x^3$ is negative, so $3x^3$ is negative, denominator $x$ is negative, so negative divided by negative is positive, and the magnitude is $3x^2$, which goes to infinity. So yes, the limit is $\infty$. Wait, but maybe I made a mistake. Wait, no, the degree of numerator is 3, denominator is 1, so 3 > 1, so the limit is $\pm\infty$. The sign: leading coefficient of numerator is 3 (positive), leading coefficient of denominator is 1 (positive). But since $x \to -\infty$, the numerator's leading term is $3x^3$ (negative, because $x^3$ is negative), denominator's leading term is $x$ (negative). So (positive * negative) / negative = positive. So the limit is $\infty$. So $\lim_{x \to -\infty} \frac{3x^3 - 10}{x + 4} = \infty$? Wait, but actually, when $x \to -\infty$, $x^3$ is negative, so $3x^3$ is negative, denominator $x$ is negative, so negative divided by negative is positive, and the magnitude is $3x^2$, which goes to infinity. So yes, the limit is $\infty$. Wait, but maybe the problem expects $-\infty$? Wait, no, let's take $x = -1$: numerator $3(-1)^3 -10 = -3 -10 = -13$, denominator $-1 +4 = 3$, so $-13/3 \approx -4.33$. $x = -10$: numerator $3(-10)^3 -10 = -3000 -10 = -3010$, denominator $-10 +4 = -6$, so $-3010 / -6 \approx 501.67$ (positive). $x = -100$: numerator $3(-100)^3 -10 = -3,000,000 -10 = -3,000,010$, denominator $-100 +4 = -96$, so $-3,000,010 / -96 \approx 31250.10$ (positive and large). So as $x$ becomes more negative (goes to $-\infty$), the value becomes positive and large, so the limit is $\infty$. So $\lim_{x \to -\infty} \frac{3x^3 - 10}{x + 4} = \infty$

Problem 16: $\lim_{x \to -\infty} \frac{2x^5 + 3x^4 - 31x}{8x^4 - 31x^2 + 12}$

Step 1: Identify degrees of numerator and denominator

Degree of numerator ($2x^5 + 3x^4 - 31x$): $5$
Degree of denominator ($8x^4 - 31x^2 + 12$): $4$

Since $5 > 4$ (degree of numerator > degree of denominator), we analyze the sign. As $x \to -\infty$:

Leading term of numerator: $2x^5$ (negative, since $x^5$ is negative when $x$ is negative)
Leading term of denominator: $8x^4$ (positive, since $x^4$ is positive when $x$ is negative)

Ratio of leading terms: $\frac{2x^5}{8x^4} = \frac{2x}{8} = \frac{x}{4}$. As $x \to -\infty$, $\frac{x}{4} \to -\infty$ (since $x$ is negative and magnitude goes to infinity).

Thus, $\lim_{x \to -\infty} \frac{2x^5 + 3x^4 - 31x}{8x^4 - 31x^2 + 12} = -\infty$

Final Answers:

$\boldsymbol{\frac{3}{4}}$
$\boldsymbol{0}$
$\boldsymbol{0}$
$\boldsymbol{\frac{7}{4}}$
$\boldsymbol{-\infty}$
$\boldsymbol{-\infty}$
$\boldsymbol{0}$

1

Sovi.AI - AI Math Tutor

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Question

Explanation:

Problem 8: \(\lim_{x \to \infty} \frac{3x^2 + 20x}{4x^2 + 9}\)

Step 1: Identify degrees of numerator and denominator

Step 2: Take ratio of leading coefficients

Problem 9: \(\lim_{x \to \infty} \frac{3x^2 + 20x}{2x^4 + 3x^3 - 29}\)

Step 1: Identify degrees of numerator and denominator

Problem 10: \(\lim_{x \to \infty} \frac{4}{x + 5}\)

Step 1: Identify degrees of numerator and denominator

Problem 11: \(\lim_{x \to \infty} \frac{7x - 9}{4x + 3}\)

Step 1: Identify degrees of numerator and denominator

Step 2: Take ratio of leading coefficients

Problem 12: \(\lim_{x \to \infty} \frac{9x^2 - 2}{6 - 29x}\)

Step 1: Identify degrees of numerator and denominator

Problem 13: \(\lim_{x \to -\infty} \frac{7x^2 - 9}{4x + 3}\)

Step 1: Identify degrees of numerator and denominator

Answer:

Problem 8: \(\lim_{x \to \infty} \frac{3x^2 + 20x}{4x^2 + 9}\)

Step 1: Identify degrees of numerator and denominator

Step 2: Take ratio of leading coefficients

Problem 9: \(\lim_{x \to \infty} \frac{3x^2 + 20x}{2x^4 + 3x^3 - 29}\)

Step 1: Identify degrees of numerator and denominator

Problem 10: \(\lim_{x \to \infty} \frac{4}{x + 5}\)

Step 1: Identify degrees of numerator and denominator

Problem 11: \(\lim_{x \to \infty} \frac{7x - 9}{4x + 3}\)

Step 1: Identify degrees of numerator and denominator

Step 2: Take ratio of leading coefficients

Problem 12: \(\lim_{x \to \infty} \frac{9x^2 - 2}{6 - 29x}\)

Step 1: Identify degrees of numerator and denominator

Problem 13: \(\lim_{x \to -\infty} \frac{7x^2 - 9}{4x + 3}\)

Step 1: Identify degrees of numerator and denominator

Problem 14: \(\lim_{x \to -\infty} \frac{5x - 9}{4x^3 + 2x + 7}\)

Step 1: Identify degrees of numerator and denominator

Problem 15: \(\lim_{x \to -\infty} \frac{3x^3 - 10}{x + 4}\)

Step 1: Identify degrees of numerator and denominator

Problem 16: \(\lim_{x \to -\infty} \frac{2x^5 + 3x^4 - 31x}{8x^4 - 31x^2 + 12}\)

Step 1: Identify degrees of numerator and denominator

Final Answers: