# Help! Limits of functions

• Dec 18th 2008, 07:44 PM
Moon Hoplite
Help! Limits of functions
Hi,

Q1.

lim (x = infinity) { (X^2 +5x - 6)/(4x^2 +x -8) }

Q2.

lim (x = infinity) { x^2/(x+5) - x }

i know how to find limits of functions, but i dont know how to do it when x is infinity, please explain,

thanks

Moon
• Dec 18th 2008, 08:13 PM
Mentia
When you have a limit that looks like 0/0 or Infinity/Infinity thats when you use L'Hopital's rule.

For the first one you can use the rule immediately. For the second one you have to get it in the form Infinity/Infinity so you get a common denominator and it turns into -5x/(x+5), and then use the rule.
• Dec 18th 2008, 08:23 PM
Soroban
Hello, Moon!

Quote:

$1)\;\;\lim_{x\to\infty}\frac{x^2 +5x - 6}{4x^2 +x -8}$
Divide top and bottom by $x^2$

$\lim_{x\to\infty}\frac{\dfrac{x^2}{x^2} + \dfrac{5x}{x^2} - \dfrac{6}{x}}{\dfrac{4x^2}{x^2} + \dfrac{x}{x^2} - \dfrac{8}{x^2}} \;=\; \lim_{x\to\infty}\frac{1 + \dfrac{5}{x} - \dfrac{6}{x^2}}{4 + \dfrac{1}{x} - \dfrac{8}{x^2}} \;=\;\frac{1+0-0}{4+0-0} \;=\;\frac{1}{4}$

Quote:

$2)\;\;\lim_{x\to\infty}\left(\frac{x^2}{x+5} - x\right)$
Combine the fractions: . $\lim_{x\to\infty}\left(\frac{x^2-x(x+5)}{x+5}\right) \;=\;\lim_{x\to\infty}\frac{-5x}{x+5}$

Divide top and bottom by $x$

. . $\lim_{x\to\infty}\frac{\dfrac{-5x}{x}}{\dfrac{x}{x} + \dfrac{5}{x}} \;=\;\lim_{x\to\infty}\frac{-5}{1 + \dfrac{5}{x}} \;=\;\frac{-5}{1+0} \;=\;-5$

• Dec 18th 2008, 08:41 PM
Chris L T521
See this post, and read my generalization for limits of rational expressions (note this is just for $\infty$. You can make slight modifications for it as it goes to $-\infty$).