Question

lesson 8.7 - limits at infinity
quicknotes
lim f(x)=9 → horizontal asymptote at y = 9
x→9
lim f(x)=9 → vertical at x = 6
x→6
to find end behavior compare numerator denominator
degree to number→ degree of denominator→ no horizontal
asymptote, degree of number degree of denominator
→ h.a at y = 0
check your understanding
use the graph of f(x) to answer questions 1 - 3.

1. evaluate each limit:

a. lim f(x)
x→4
b. lim f(x)
x→0-
c. lim f(x)
x→0+
d. lim f(x)
x→-∞
e. lim f(x)
x→3-

1. list all vertical asymptotes.
2. list all horizontal asymptotes.
3. for which of the following functions is lim f(x)=0?

x→∞
i. f(x)=\frac{3x - 5}{e^x}
ii. f(x)=\frac{7x^2+4x^3}{5x^3-6x^2+2x - 1}
iii. f(x)=\frac{(x - 2)}{(x + 4)(x - 5)}
a) i only
b) ii only
c) ii and iii only
d) i and iii only
e) i, ii, and iii

1. write the equation of a function g(x) where lim g(x)=\frac{4}{5} and lim g(x)=-∞

x→∞ x→-3+
math medic

Explanation:

Step1: Analyze limits from graph for 1.a - e

For limits as $x$ approaches a value, check the graph behavior near that $x -$value. For limits as $x\to\pm\infty$, check the end - behavior.

Step2: Identify vertical asymptotes for 2

Vertical asymptotes occur where the function approaches $\pm\infty$. Look for $x$ - values where the graph has a break and goes to infinity.

Step3: Identify horizontal asymptotes for 3

Compare the degrees of the numerator and denominator of the rational function (if applicable) or look at the end - behavior of the graph. If the degree of the numerator is less than the degree of the denominator, $y = 0$ is a horizontal asymptote. If the degrees are equal, the horizontal asymptote is $y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$.

Step4: Evaluate limits for 4

For $f(x)=\frac{3x - 5}{e^{x}}$, use L'Hopital's rule or the fact that exponential growth ($e^{x}$) outpaces linear growth ($3x-5$) as $x\to\infty$. For $f(x)=\frac{7x^{2}+4x^{3}}{5x^{3}-6x^{2}+2x - 1}$, divide numerator and denominator by $x^{3}$ and find the limit as $x\to\infty$. For $f(x)=\frac{x - 2}{(x + 4)(x - 5)}=\frac{x - 2}{x^{2}-x - 20}$, divide numerator and denominator by $x^{2}$ and find the limit as $x\to\infty$.

Step5: Construct function for 5

To have $\lim_{x\to\infty}g(x)=\frac{4}{5}$, the function should be a rational function with the ratio of leading coefficients $\frac{4}{5}$ for numerator and denominator of the same degree. To have $\lim_{x\to - 3^{+}}g(x)=-\infty$, there should be a factor of $(x + 3)$ in the denominator. A possible function is $g(x)=\frac{4x+12}{5x + 15}$.

Answer:

1. (Answers depend on the graph, not provided here in detail. For example, if as $x\to4$ the graph approaches a value $L$, then $\lim_{x\to4}f(x)=L$ etc.)
2. (Answers depend on the graph, vertical asymptotes are $x$ - values where the graph goes to $\pm\infty$)
3. (Answers depend on the graph or function analysis as described above)
4. E. I, II, and III

• For $I$: $\lim_{x\to\infty}\frac{3x - 5}{e^{x}}$, using L'Hopital's rule (since $\lim_{x\to\infty}\frac{3x - 5}{e^{x}}$ is in $\frac{\infty}{\infty}$ form), $\lim_{x\to\infty}\frac{3}{e^{x}} = 0$.
• For $II$: $\lim_{x\to\infty}\frac{7x^{2}+4x^{3}}{5x^{3}-6x^{2}+2x - 1}=\lim_{x\to\infty}\frac{\frac{7}{x}+4}{5-\frac{6}{x}+\frac{2}{x^{2}}-\frac{1}{x^{3}}}=\frac{4}{5}$.
• For $III$: $\lim_{x\to\infty}\frac{x - 2}{(x + 4)(x - 5)}=\lim_{x\to\infty}\frac{\frac{1}{x}-\frac{2}{x^{2}}}{1-\frac{1}{x}-\frac{20}{x^{2}}}=0$.

1. $g(x)=\frac{4x + 12}{5x+15}$ (or other equivalent functions)