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
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.
Hello, Moon!
Divide top and bottom by $\displaystyle x^2$$\displaystyle 1)\;\;\lim_{x\to\infty}\frac{x^2 +5x - 6}{4x^2 +x -8}$
$\displaystyle \lim_{x\to\infty}\frac{\dfrac{x^2}{x^2} + \dfrac{5x}{x^2} - \dfrac{6}{x}}{\dfrac{4x^2}{x^2} + \dfrac{x}{x^2} - \dfrac{8}{x^2}} \;=\; \lim_{x\to\infty}\frac{1 + \dfrac{5}{x} - \dfrac{6}{x^2}}{4 + \dfrac{1}{x} - \dfrac{8}{x^2}} \;=\;\frac{1+0-0}{4+0-0} \;=\;\frac{1}{4}$
Combine the fractions: .$\displaystyle \lim_{x\to\infty}\left(\frac{x^2-x(x+5)}{x+5}\right) \;=\;\lim_{x\to\infty}\frac{-5x}{x+5}$$\displaystyle 2)\;\;\lim_{x\to\infty}\left(\frac{x^2}{x+5} - x\right)$
Divide top and bottom by $\displaystyle x$
. . $\displaystyle \lim_{x\to\infty}\frac{\dfrac{-5x}{x}}{\dfrac{x}{x} + \dfrac{5}{x}} \;=\;\lim_{x\to\infty}\frac{-5}{1 + \dfrac{5}{x}} \;=\;\frac{-5}{1+0} \;=\;-5$