1.) \lim_{x \to -\infty} \frac{3x^5 - 6x^4 + 7x^3 - 12x^2 + 5x}{9x^5 + 7x^4 - 6x^3 - 3x^2}
2.) \lim_{x \to \infty} \frac{-4x^3 - 7x^2 + 8x - 3}{7x^4 - 3x^3 + 5x^2 - 7x}
3.) \lim_{x \to \infty} \frac{-10x^3 - 5x^2 + 2x - 4}{4x^3 - 8x^2 + 7x + 3}
4.) \lim_{x \to -\infty} \frac{3x^4 + 9x^3 - 2x^2 + 7x}{-6x^2 - 7x + 5}

Question

Accepted Answer

### Problem 1: $\lim\limits_{x 	o -\infty} \frac{3x^5 - 6x^4 + 7x^3 - 12x^2 + 5x}{9x^5 + 7x^4 - 6x^3 - 3x^2}$
# Explanation:
## Step 1: Divide numerator and denominator by $x^5$ (highest power in denominator)
For large $|x|$, the terms with the highest power dominate. So we divide each term in numerator and denominator by $x^5$.
$$
\begin{align*}
\lim\limits_{x 	o -\infty} \frac{\frac{3x^5}{x^5} - \frac{6x^4}{x^5} + \frac{7x^3}{x^5} - \frac{12x^2}{x^5} + \frac{5x}{x^5}}{\frac{9x^5}{x^5} + \frac{7x^4}{x^5} - \frac{6x^3}{x^5} - \frac{3x^2}{x^5}}&=\lim\limits_{x 	o -\infty} \frac{3 - \frac{6}{x} + \frac{7}{x^2} - \frac{12}{x^3} + \frac{5}{x^4}}{9 + \frac{7}{x} - \frac{6}{x^2} - \frac{3}{x^3}}
\end{align*}
$$
## Step 2: Evaluate limit as $x 	o -\infty$
As $x 	o -\infty$, terms with $\frac{1}{x^n}$ (where $n>0$) approach 0. So we substitute the limits of each term.
$$
\frac{3 - 0 + 0 - 0 + 0}{9 + 0 - 0 - 0}=\frac{3}{9}=\frac{1}{3}
$$
# Answer: $\frac{1}{3}$

### Problem 2: $\lim\limits_{x 	o \infty} \frac{-4x^3 - 7x^2 + 8x - 3}{7x^4 - 3x^3 + 5x^2 - 7x}$
# Explanation:
## Step 1: Divide numerator and denominator by $x^4$ (highest power in denominator)
Divide each term in numerator and denominator by $x^4$.
$$
\begin{align*}
\lim\limits_{x 	o \infty} \frac{\frac{-4x^3}{x^4} - \frac{7x^2}{x^4} + \frac{8x}{x^4} - \frac{3}{x^4}}{\frac{7x^4}{x^4} - \frac{3x^3}{x^4} + \frac{5x^2}{x^4} - \frac{7x}{x^4}}&=\lim\limits_{x 	o \infty} \frac{-\frac{4}{x} - \frac{7}{x^2} + \frac{8}{x^3} - \frac{3}{x^4}}{7 - \frac{3}{x} + \frac{5}{x^2} - \frac{7}{x^3}}
\end{align*}
$$
## Step 2: Evaluate limit as $x 	o \infty$
As $x 	o \infty$, terms with $\frac{1}{x^n}$ (where $n>0$) approach 0. Substitute the limits.
$$
\frac{0 - 0 + 0 - 0}{7 - 0 + 0 - 0}=0
$$
# Answer: $0$

### Problem 3: $\lim\limits_{x 	o \infty} \frac{-10x^3 - 5x^2 + 2x - 4}{4x^3 - 8x^2 + 7x + 3}$
# Explanation:
## Step 1: Divide numerator and denominator by $x^3$ (highest power in numerator and denominator)
Divide each term in numerator and denominator by $x^3$.
$$
\begin{align*}
\lim\limits_{x 	o \infty} \frac{\frac{-10x^3}{x^3} - \frac{5x^2}{x^3} + \frac{2x}{x^3} - \frac{4}{x^3}}{\frac{4x^3}{x^3} - \frac{8x^2}{x^3} + \frac{7x}{x^3} + \frac{3}{x^3}}&=\lim\limits_{x 	o \infty} \frac{-10 - \frac{5}{x} + \frac{2}{x^2} - \frac{4}{x^3}}{4 - \frac{8}{x} + \frac{7}{x^2} + \frac{3}{x^3}}
\end{align*}
$$
## Step 2: Evaluate limit as $x 	o \infty$
As $x 	o \infty$, terms with $\frac{1}{x^n}$ (where $n>0$) approach 0. Substitute the limits.
$$
\frac{-10 - 0 + 0 - 0}{4 - 0 + 0 + 0}=-\frac{10}{4}=-\frac{5}{2}
$$
# Answer: $-\frac{5}{2}$

### Problem 4: $\lim\limits_{x 	o -\infty} \frac{3x^4 + 9x^3 - 2x^2 + 7x}{-6x^2 - 7x + 5}$
# Explanation:
## Step 1: Divide numerator and denominator by $x^4$ (highest power in numerator)
Divide each term in numerator and denominator by $x^4$.
$$
\begin{align*}
\lim\limits_{x 	o -\infty} \frac{\frac{3x^4}{x^4} + \frac{9x^3}{x^4} - \frac{2x^2}{x^4} + \frac{7x}{x^4}}{\frac{-6x^2}{x^4} - \frac{7x}{x^4} + \frac{5}{x^4}}&=\lim\limits_{x 	o -\infty} \frac{3 + \frac{9}{x} - \frac{2}{x^2} + \frac{7}{x^3}}{-\frac{6}{x^2} - \frac{7}{x^3} + \frac{5}{x^4}}
\end{align*}
$$
## Step 2: Analyze the limit
As $x 	o -\infty$, the numerator approaches $3 + 0 - 0 + 0 = 3$, and the denominator approaches $0 - 0 + 0 = 0$. But we also note the sign: when $x$ is a large negative number, $x^4$ is positive, $x^3$ is negative, $x^2$ is positive, $x$ is negative. Let's check the leading terms for large $|x|$ (we can also divide numerator and denominator by $x^2$ for easier analysis, since the denominator's highest power is $x^2$ and numerator's is $x^4$). Let's redo step 1 by dividing by $x^2$:
$$
\begin{align*}
\lim\limits_{x 	o -\infty} \frac{\frac{3x^4}{x^2} + \frac{9x^3}{x^2} - \frac{2x^2}{x^2} + \frac{7x}{x^2}}{\frac{-6x^2}{x^2} - \frac{7x}{x^2} + \frac{5}{x^2}}&=\lim\limits_{x 	o -\infty} \frac{3x^2 + 9x - 2 + \frac{7}{x}}{-6 - \frac{7}{x} + \frac{5}{x^2}}
\end{align*}
$$
As $x 	o -\infty$, $3x^2$ (positive, since square) dominates the numerator, and the denominator approaches $-6$. So the numerator goes to $+\infty$ (since $x^2$ is positive and large) and denominator approaches $-6$ (negative). So the limit is $-\infty$. Wait, but let's check the first approach. When we divide by $x^4$, numerator limit is 3, denominator limit is 0. But the denominator: $-\frac{6}{x^2} - \frac{7}{x^3} + \frac{5}{x^4}$. As $x 	o -\infty$, $x^2$ is positive, so $-\frac{6}{x^2}$ is negative, $x^3$ is negative, so $-\frac{7}{x^3}$ is positive (negative divided by negative), but as $x 	o -\infty$, $x^3$ is very negative, so $-\frac{7}{x^3}$ is a small positive, and $\frac{5}{x^4}$ is positive and small. But the dominant term in denominator when $x 	o -\infty$ (dividing by $x^2$) is better. Let's use the rule for limits of rational functions: if degree of numerator ($n$) > degree of denominator ($m$), then if $n - m$ is even and leading coefficients have same sign, limit is $+\infty$, else $-\infty$; if $n - m$ is odd, sign depends on leading coefficients and $x$ approaching $\pm\infty$. Here, numerator degree is 4, denominator degree is 2. $4 - 2 = 2$ (even). Leading coefficient of numerator: 3 (positive), leading coefficient of denominator: -6 (negative). So positive divided by negative, and even degree difference: limit is $-\infty$ (since as $x 	o -\infty$, $x^4$ is positive, $x^2$ is positive, so numerator ~ $3x^4$, denominator ~ $-6x^2$, so $\frac{3x^4}{-6x^2}=-\frac{3}{6}x^2=-\frac{1}{2}x^2$, which goes to $-\infty$ as $x 	o -\infty$ (since $x^2$ is positive and large, multiplied by negative gives $-\infty$).
# Answer: $-\infty$

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Question

Explanation:

Problem 1: $\lim\limits_{x \to -\infty} \frac{3x^5 - 6x^4 + 7x^3 - 12x^2 + 5x}{9x^5 + 7x^4 - 6x^3 - 3x^2}$

Step 1: Divide numerator and denominator by $x^5$ (highest power in denominator)

Step 2: Evaluate limit as $x \to -\infty$

Step 1: Divide numerator and denominator by $x^4$ (highest power in denominator)

Step 2: Evaluate limit as $x \to \infty$

Step 1: Divide numerator and denominator by $x^3$ (highest power in numerator and denominator)

Step 2: Evaluate limit as $x \to \infty$

Answer:

Problem 2: $\lim\limits_{x \to \infty} \frac{-4x^3 - 7x^2 + 8x - 3}{7x^4 - 3x^3 + 5x^2 - 7x}$