6. evaluate the following limits using calculus methods and draw conclusions about what they mean about the function.
b) $limlimits_{x \to -1^-} \frac{-1}{x^3 + x^2}$
c) $limlimits_{x \to 3^+} \frac{x^2 + 9}{x - 3}$
d) $limlimits_{x \to 3} \frac{x^2 - 9}{x - 3}$
e) $limlimits_{x \to \infty} \frac{3x}{5x + 10}$
f) $limlimits_{x \to -\infty} \frac{2x^4 + 5x^3 - 1}{3x^2 - 2x + 8}$
g) $limlimits_{x \to \infty} \frac{5}{x^3 + 1}$

Question

Accepted Answer

### Part (b)
# Explanation:
## Step 1: Simplify the denominator
Factor the denominator $x^3 + x^2=x^2(x + 1)$
So the function becomes $\frac{-1}{x^2(x + 1)}$

## Step 2: Analyze the limit as $x
ightarrow - 1^{-}$
As $x
ightarrow - 1^{-}$, $x+1
ightarrow0^{-}$ (since $x$ is slightly less than - 1) and $x^2>0$ (because square of any non - zero real number is positive).
So the denominator $x^2(x + 1)$ is the product of a positive number ($x^2$) and a negative number approaching 0 ($x + 1
ightarrow0^{-}$). So $x^2(x + 1)
ightarrow0^{-}$
The numerator is $- 1$ (a constant).
Then $\lim_{x
ightarrow - 1^{-}}\frac{-1}{x^3+x^2}=\lim_{x
ightarrow - 1^{-}}\frac{-1}{x^2(x + 1)}$
Since the numerator is $-1$ and the denominator approaches $0^{-}$, the limit is $+\infty$ (because $\frac{-1}{0^{-}}=\frac{1}{0^{+}}=+\infty$)

### Part (c)
# Explanation:
## Step 1: Analyze the numerator and denominator as $x
ightarrow3^{+}$
As $x
ightarrow3^{+}$, the numerator $x^{2}+9=3^{2}+9 = 18>0$
The denominator $x - 3
ightarrow0^{+}$ (since $x$ is slightly greater than 3)

## Step 2: Evaluate the limit
We have a positive numerator and a denominator that approaches $0^{+}$. So $\lim_{x
ightarrow3^{+}}\frac{x^{2}+9}{x - 3}=+\infty$

### Part (d)
# Explanation:
## Step 1: Factor the numerator
We know that $x^{2}-9=(x - 3)(x + 3)$ (using the difference of squares formula $a^{2}-b^{2}=(a - b)(a + b)$ where $a=x$ and $b = 3$)
So the function $\frac{x^{2}-9}{x - 3}=\frac{(x - 3)(x + 3)}{x - 3}$ for $x
eq3$

## Step 2: Cancel out the common factor
Cancel out the common factor $(x - 3)$ (since $x
ightarrow3$ but $x
eq3$ when taking the limit), we get $y=x + 3$

## Step 3: Evaluate the limit
Now, $\lim_{x
ightarrow3}\frac{x^{2}-9}{x - 3}=\lim_{x
ightarrow3}(x + 3)$
Substitute $x = 3$ into $x+3$, we get $3 + 3=6$

### Part (e)
# Explanation:
## Step 1: Divide numerator and denominator by $x$
Divide the numerator $3x$ and the denominator $5x + 10$ by $x$ (since $x
ightarrow\infty$, $x
eq0$)
We have $\frac{3x}{5x + 10}=\frac{\frac{3x}{x}}{\frac{5x}{x}+\frac{10}{x}}=\frac{3}{5+\frac{10}{x}}$

## Step 2: Evaluate the limit as $x
ightarrow\infty$
As $x
ightarrow\infty$, $\frac{10}{x}
ightarrow0$
So $\lim_{x
ightarrow\infty}\frac{3}{5+\frac{10}{x}}=\frac{3}{5 + 0}=\frac{3}{5}$

### Part (f)
# Explanation:
## Step 1: Divide numerator and denominator by the highest power of $x$ in the denominator
The highest power of $x$ in the denominator $3x^{2}-2x + 8$ is $x^{2}$. Divide numerator $2x^{4}+5x^{3}-1$ and denominator $3x^{2}-2x + 8$ by $x^{2}$
$\frac{2x^{4}+5x^{3}-1}{3x^{2}-2x + 8}=\frac{\frac{2x^{4}}{x^{2}}+\frac{5x^{3}}{x^{2}}-\frac{1}{x^{2}}}{\frac{3x^{2}}{x^{2}}-\frac{2x}{x^{2}}+\frac{8}{x^{2}}}=\frac{2x^{2}+5x-\frac{1}{x^{2}}}{3-\frac{2}{x}+\frac{8}{x^{2}}}$

## Step 2: Evaluate the limit as $x
ightarrow-\infty$
As $x
ightarrow-\infty$, $x^{2}
ightarrow+\infty$, $x
ightarrow-\infty$, $\frac{1}{x^{2}}
ightarrow0$, $\frac{2}{x}
ightarrow0$, $\frac{8}{x^{2}}
ightarrow0$
The numerator $2x^{2}+5x-\frac{1}{x^{2}}$ behaves like $2x^{2}$ (since $x^{2}$ term dominates) and as $x
ightarrow-\infty$, $2x^{2}
ightarrow+\infty$
The denominator $3-\frac{2}{x}+\frac{8}{x^{2}}
ightarrow3$
So $\lim_{x
ightarrow-\infty}\frac{2x^{4}+5x^{3}-1}{3x^{2}-2x + 8}=-\infty$ (because the numerator goes to $-\infty$ when considering the sign: $x
ightarrow-\infty$, $x^{2}$ is positive, but the term $5x$ with $x
ightarrow-\infty$ makes the leading term of the numerator $2x^{2}+5x$ go to $-\infty$ as $x
ightarrow-\infty$ (since for large negative $x$, $5x$ dominates $2x^{2}$ in terms of sign? Wait, no: Wait, $x
ightarrow-\infty$, let $x=-t$ where $t
ightarrow+\infty$
Then $2x^{2}+5x=2t^{2}-5t$, as $t
ightarrow+\infty$, $2t^{2}-5t
ightarrow+\infty$ (since $t^{2}$ term dominates). Wait, I made a mistake earlier.
Let's re - do step 2:
$\lim_{x
ightarrow-\infty}\frac{2x^{4}+5x^{3}-1}{3x^{2}-2x + 8}$, divide numerator and denominator by $x^{4}$ (highest power in numerator)
$\frac{2x^{4}+5x^{3}-1}{3x^{2}-2x + 8}=\frac{2+\frac{5}{x}-\frac{1}{x^{4}}}{\frac{3}{x^{2}}-\frac{2}{x^{3}}+\frac{8}{x^{4}}}$
As $x
ightarrow-\infty$, $\frac{5}{x}
ightarrow0$, $\frac{1}{x^{4}}
ightarrow0$, $\frac{3}{x^{2}}
ightarrow0$, $\frac{2}{x^{3}}
ightarrow0$, $\frac{8}{x^{4}}
ightarrow0$
So the limit is $\frac{2+0 - 0}{0-0 + 0}=\infty$. But since the leading coefficient of numerator is positive and the degree of numerator ($4$) is greater than degree of denominator ($2$), as $x
ightarrow-\infty$, $x^{4}$ is positive, so the limit is $+\infty$? Wait, no, when $x
ightarrow-\infty$, $x^{4}=(x^{2})^{2}>0$, $x^{3}<0$
Wait, let's use the rule for limits at infinity of rational functions: If $f(x)=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}$, $a_n
eq0$, $b_m
eq0$
- If $n>m$, $\lim_{x
ightarrow\pm\infty}f(x)=\pm\infty$ (the sign depends on the sign of $a_n$ and the parity of $n - m$ and the sign of $x$ for odd powers). Here $n = 4$, $m = 2$, $n-m = 2$ (even), $a_n=2>0$, so as $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{2x^{4}+5x^{3}-1}{3x^{2}-2x + 8}=+\infty$

### Part (g)
# Explanation:
## Step 1: Analyze the numerator and denominator as $x
ightarrow\infty$
The numerator is $5$ (a constant)
As $x
ightarrow\infty$, the denominator $x^{3}+1
ightarrow\infty$ (since $x^{3}$ dominates as $x
ightarrow\infty$)

## Step 2: Evaluate the limit
We know that $\lim_{x
ightarrow\infty}\frac{c}{x^n}=0$ for $c\in\mathbb{R}$ and $n>0$
So $\lim_{x
ightarrow\infty}\frac{5}{x^{3}+1}=0$ (because $\frac{5}{x^{3}+1}$ behaves like $\frac{5}{x^{3}}$ as $x
ightarrow\infty$ and $\lim_{x
ightarrow\infty}\frac{5}{x^{3}} = 0$)

# Answers:
b) $\boldsymbol{+\infty}$

c) $\boldsymbol{+\infty}$

d) $\boldsymbol{6}$

e) $\boldsymbol{\frac{3}{5}}$

f) $\boldsymbol{+\infty}$

g) $\boldsymbol{0}$

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QUESTION IMAGE

Question

Explanation:

Part (b)

Step 1: Simplify the denominator

Step 2: Analyze the limit as \(x

Part (c)

Step 1: Analyze the numerator and denominator as \(x

Step 2: Evaluate the limit

Part (d)

Step 1: Factor the numerator

Step 2: Cancel out the common factor

Step 3: Evaluate the limit

Part (e)

Answer: