# Thread: Finding limits at infinity - help on a couple problems

1. ## Finding limits at infinity - help on a couple problems

I managed to get the first couple questions in the section right..but as most of you know they get harder the farther you go. Currently these are the two I am stuck on. If somebody could show me how to solve them I can hopefully get the hang of the rest of them.

Decide whether the limits in the following problems exist. If a limit exist, find its value. The set of numbers should be next to the limit x approaching infinity but it came out sloppy so I just put them below.

Problem 1
lim
x-->infinity

x^2 + 6x + 8
x^3 + 2x + 1

Problem 2
lim
x--> - infinity

( 9/x^4 + 1/x^2 - 3)

Their are parentheses surrounding the entire problem except the limit(In other words from the 9/x^4 to the -3) I didn't put them as a fraction with an underline like problem 1 because they are seperate and it came out too sloppy, but you get the idea, it's 9 over x^4 plus 1 over x^2 minus 3

Thanks in advance for the help!

2. Originally Posted by Mathamateur

Problem 1
lim
x-->infinity

x^2 + 6x + 8
x^3 + 2x + 1
By dividing throw by the largest exponent $x^3$ these functions agree for sufficiently large $x$.

Thus,
$\lim_{x\to \infty} \frac{\frac{1}{x}+\frac{6}{x^2}+\frac{8}{x^3}}{1+\ frac{2}{x^2}+\frac{1}{x^3}}=\frac{0+0+0}{1+0+0}=0$

Problem 2
lim
x--> - infinity

( 9/x^4 + 1/x^2 - 3)
!
That is,
$0+0-3=-3$

3. Originally Posted by Mathamateur
Problem 1
lim
x-->infinity

x^2 + 6x + 8
x^3 + 2x + 1
$\lim_{x \to \infty} \frac{x^2+6x+8}{x^3+2x+1}$

Divide the numerator and denominator by $\frac{1}{x^2}$:

= $\lim_{x \to \infty}\frac{1 + \frac{6}{x} + \frac{8}{x^2}}{x + \frac{2}{x} + \frac{1}{x^2}}$

Taking the limit of the numerator denominator we see that $\lim_{x \to \infty}\frac{1}{x}$ and $\lim_{x \to \infty}\frac{1}{x^2}$ are both 0, so

$\lim_{x \to \infty} \frac{x^2+6x+8}{x^3+2x+1} \to \lim_{x \to \infty}\frac{1}{x} \to 0$

-Dan

4. So does the limit not exist, or is the limit zero?

5. Why would you be thinking it wouldn't exist?

The limit in the first is 0.
The limit in the second is -3.

-Dan

6. Originally Posted by Mathamateur
So does the limit not exist, or is the limit zero?
Whenever a limit is equal to a number it exists.
Otherwise, it does not exist.