1.) $\lim\limits_{x \to -\infty} \frac{3x^5 - 6x^4 + 7x^3 - 12x^2 + 5x}{9x^5 + 7x^4 - 6x^3 - 3x^2}$
2.) $\lim\limits_{x \to \infty} \frac{-4x^3 - 7x^2 + 8x - 3}{7x^4 - 3x^3 + 5x^2 - 7x}$
3.) $\lim\limits_{x \to \infty} \frac{-10x^3 - 5x^2 + 2x - 4}{4x^3 - 8x^2 + 7x + 3}$
4.) $\lim\limits_{x \to -\infty} \frac{3x^4 + 9x^3 - 2x^2 + 7x}{-6x^2 - 7x + 5}$
act #3

Question

Accepted Answer

### Problem 1: $\boldsymbol{\lim_{x 	o -\infty} \frac{3x^5 - 6x^4 + 7x^3 - 12x^2 + 5x}{9x^5 + 7x^4 - 6x^3 - 3x^2}}$
# Explanation:
## Step1: 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 $x^5$:
$$
\begin{align*}
\lim_{x 	o -\infty} \frac{\frac{3x^5}{x^5} - \frac{6x^4}{x^5} + \frac{7x^3}{x^5} - \frac{12x^2}{x^5} + \frac{5x}{x^5}}{\frac{9x^5}{x^5} + \frac{7x^4}{x^5} - \frac{6x^3}{x^5} - \frac{3x^2}{x^5}}&=\lim_{x 	o -\infty} \frac{3 - \frac{6}{x} + \frac{7}{x^2} - \frac{12}{x^3} + \frac{5}{x^4}}{9 + \frac{7}{x} - \frac{6}{x^2} - \frac{3}{x^3}}
\end{align*}
$$
## Step2: Evaluate limit as $x 	o -\infty$
As $x 	o -\infty$, terms with $\frac{1}{x^n}$ (where $n>0$) approach 0. So:
$$
\frac{3 - 0 + 0 - 0 + 0}{9 + 0 - 0 - 0}=\frac{3}{9}=\frac{1}{3}
$$
# Answer: $\boldsymbol{\frac{1}{3}}$

### Problem 2: $\boldsymbol{\lim_{x 	o \infty} \frac{-4x^3 - 7x^2 + 8x - 3}{7x^4 - 3x^3 + 5x^2 - 7x}}$
# Explanation:
## Step1: Divide numerator and denominator by $x^4$ (highest power in denominator)
Divide each term in numerator and denominator by $x^4$:
$$
\begin{align*}
\lim_{x 	o \infty} \frac{\frac{-4x^3}{x^4} - \frac{7x^2}{x^4} + \frac{8x}{x^4} - \frac{3}{x^4}}{\frac{7x^4}{x^4} - \frac{3x^3}{x^4} + \frac{5x^2}{x^4} - \frac{7x}{x^4}}&=\lim_{x 	o \infty} \frac{-\frac{4}{x} - \frac{7}{x^2} + \frac{8}{x^3} - \frac{3}{x^4}}{7 - \frac{3}{x} + \frac{5}{x^2} - \frac{7}{x^3}}
\end{align*}
$$
## Step2: Evaluate limit as $x 	o \infty$
As $x 	o \infty$, terms with $\frac{1}{x^n}$ (where $n>0$) approach 0. So:
$$
\frac{0 - 0 + 0 - 0}{7 - 0 + 0 - 0}=0
$$
# Answer: $\boldsymbol{0}$

### Problem 3: $\boldsymbol{\lim_{x 	o \infty} \frac{-10x^3 - 5x^2 + 2x - 4}{4x^3 - 8x^2 + 7x + 3}}$
# Explanation:
## Step1: Divide numerator and denominator by $x^3$ (highest power in numerator/denominator)
Divide each term in numerator and denominator by $x^3$:
$$
\begin{align*}
\lim_{x 	o \infty} \frac{\frac{-10x^3}{x^3} - \frac{5x^2}{x^3} + \frac{2x}{x^3} - \frac{4}{x^3}}{\frac{4x^3}{x^3} - \frac{8x^2}{x^3} + \frac{7x}{x^3} + \frac{3}{x^3}}&=\lim_{x 	o \infty} \frac{-10 - \frac{5}{x} + \frac{2}{x^2} - \frac{4}{x^3}}{4 - \frac{8}{x} + \frac{7}{x^2} + \frac{3}{x^3}}
\end{align*}
$$
## Step2: Evaluate limit as $x 	o \infty$
As $x 	o \infty$, terms with $\frac{1}{x^n}$ (where $n>0$) approach 0. So:
$$
\frac{-10 - 0 + 0 - 0}{4 - 0 + 0 + 0}=-\frac{10}{4}=-\frac{5}{2}
$$
# Answer: $\boldsymbol{-\frac{5}{2}}$

### Problem 4: $\boldsymbol{\lim_{x 	o -\infty} \frac{3x^4 + 9x^3 - 2x^2 + 7x}{-6x^2 - 7x + 5}}$
# Explanation:
## Step1: Divide numerator and denominator by $x^4$ (highest power in numerator)
Divide each term in numerator and denominator by $x^4$:
$$
\begin{align*}
\lim_{x 	o -\infty} \frac{\frac{3x^4}{x^4} + \frac{9x^3}{x^4} - \frac{2x^2}{x^4} + \frac{7x}{x^4}}{\frac{-6x^2}{x^4} - \frac{7x}{x^4} + \frac{5}{x^4}}&=\lim_{x 	o -\infty} \frac{3 + \frac{9}{x} - \frac{2}{x^2} + \frac{7}{x^3}}{-\frac{6}{x^2} - \frac{7}{x^3} + \frac{5}{x^4}}
\end{align*}
$$
## Step2: Analyze behavior as $x 	o -\infty$
As $x 	o -\infty$, numerator approaches $3 + 0 - 0 + 0 = 3$, and denominator approaches $0 - 0 + 0 = 0$. But we also check the sign: for large negative $x$, denominator $-\frac{6}{x^2}$ (since $x^2$ is positive, $-\frac{6}{x^2}$ is negative) and other terms are negligible. So denominator approaches 0 from the negative side (since $-\frac{6}{x^2}<0$ for all $x
eq0$). So the limit is $-\infty$ (since numerator is positive and denominator approaches 0 from negative side, positive/negative = negative, and magnitude goes to infinity). Wait, wait, let's re - check the degrees. The degree of numerator is 4, degree of denominator is 2. When degree of numerator > degree of denominator, and we have $x	o -\infty$, for rational functions, if the leading coefficient of numerator is positive and leading coefficient of denominator is negative (wait, let's re - express the division by highest power of denominator? Wait, maybe I made a mistake in step 1. Let's correct:

The degree of numerator is 4, degree of denominator is 2. So when we take limit as $x	o \pm\infty$, for a rational function $\frac{f(x)}{g(x)}$ where $\deg(f) = n$, $\deg(g)=m$. If $n>m$, then $\lim_{x	o \pm\infty}\frac{f(x)}{g(x)}=\pm\infty$ (or $-\infty$) depending on the leading coefficients and the sign of $x$.

Leading term of numerator: $3x^4$ (positive for all $x$), leading term of denominator: $-6x^2$ (negative for all $x$). So $\frac{3x^4}{-6x^2}=-\frac{3}{6}x^{2}=-\frac{1}{2}x^{2}$. As $x	o -\infty$, $x^{2}	o\infty$, so $-\frac{1}{2}x^{2}	o -\infty$. So the limit is $-\infty$.

Wait, my first step was wrong because I divided by $x^4$ (numerator's highest power) but the denominator's highest power is $x^2$. Let's do it correctly: divide numerator and denominator by $x^2$ (highest power in denominator):

$$
\begin{align*}
\lim_{x 	o -\infty} \frac{\frac{3x^4}{x^2} + \frac{9x^3}{x^2} - \frac{2x^2}{x^2} + \frac{7x}{x^2}}{\frac{-6x^2}{x^2} - \frac{7x}{x^2} + \frac{5}{x^2}}&=\lim_{x 	o -\infty} \frac{3x^2 + 9x - 2 + \frac{7}{x}}{-6 - \frac{7}{x} + \frac{5}{x^2}}
\end{align*}
$$

Now, as $x	o -\infty$, $3x^2	o\infty$ (since $x^2$ is positive and large), $9x	o -\infty$, but $3x^2$ dominates $9x$ (because degree of $x^2$ is higher than degree of $x$). The numerator: $3x^2 + 9x - 2+\frac{7}{x}\approx 3x^2$ (for large $|x|$), denominator: $-6-\frac{7}{x}+\frac{5}{x^2}\approx - 6$ (for large $|x|$). So $\frac{3x^2}{-6}=-\frac{1}{2}x^2$. As $x	o -\infty$, $x^2	o\infty$, so $-\frac{1}{2}x^2	o -\infty$.

# Answer: $\boldsymbol{-\infty}$

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QUESTION IMAGE

Question

Explanation:

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

Step1: Divide numerator and denominator by $x^5$ (highest power in denominator)

Step2: Evaluate limit as $x \to -\infty$

Step1: Divide numerator and denominator by $x^4$ (highest power in denominator)

Step2: Evaluate limit as $x \to \infty$

Step1: Divide numerator and denominator by $x^3$ (highest power in numerator/denominator)

Step2: Evaluate limit as $x \to \infty$

Answer:

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