# Evaluating a limit to infinity

• Nov 11th 2007, 05:31 PM
simsima_1
Evaluating a limit to infinity
lim x/(sqrt(x^2+1000))
x-> + inf

the answer is not as important as its how its actually done. Thanks in advance.
• Nov 11th 2007, 05:42 PM
Jhevon
Quote:

Originally Posted by simsima_1
lim x/(sqrt(x^2+1000))
x-> + inf

the answer is not as important as its how its actually done. I think L'Hopitals rule is used but i'm not so sure about the concept. Thanks in advance.

ou can try L'Hopital's but i think it will be a pain here. (we can use L'Hopital's if when we take the limit our function goes to $\frac 00$ or $\frac {\infty}{\infty}$)

instead you should realize that as $x \to \infty$ the + 1000 does not matter. thus

$\lim_{x \to \pm \infty} \frac x{\sqrt{x^2 + 1000}} = \lim_{x \to \pm \infty}\frac x{\sqrt{x^2}} = \lim_{x \to \pm \infty} \frac x{|x|} = \left \{ \begin{array}{lr} 1 & \mbox{ as } x \to \infty \\ -1 & \mbox{ as } x \to - \infty\end{array} \right.$
• Nov 12th 2007, 12:15 AM
CaptainBlack
Quote:

Originally Posted by simsima_1
lim x/(sqrt(x^2+1000))
x-> + inf

the answer is not as important as its how its actually done. Thanks in advance.

Divide top and bottom by $x$ to get:

$\lim_{x \to \infty} x/(\sqrt{x^2+1000})=\lim_{x \to \infty} \rm{sgn}(x)/(\sqrt{1+1000/x^2})=1$

RonL