Question

I am looking to use squeeze theorem to find the horizontal asymptote(s) of 1/(sqrt(x^2+1) +x). I can figure it out by dividing by x but not with squeeze theorem.

Is this to advanced for this site?

Answer 1

I hope it's not. Unfortunately, I've only touched on squeeze theorem briefly.

Answer 2

All I know is that vertical asymptotes stop the squeeze theorem working for things like tanx

Answer 3

\[\frac{1}{\sqrt{x^2+1} +x}*\frac{-\sqrt{x^2+1} +x}{-\sqrt{x^2+1} +x}\]

\[\frac{-\sqrt{x^2+1} +x}{-(x^2+1) +x^2}\]

\[\lim_{x->\inf}\sqrt{x^2+1} -x\]
maybe

Answer 4

I actually started with what you ended with. I want to use Squeeze Theorem to prove that the limit is 0

f(x) <= g(x) <= h(x) then if lim f(x) = lim g(x) = L then the lim h(x) = L too.... I think I am going to let this one go. It has taken up to much of my day. I think this site is going to be very helpful though :)

Answer 5

:) at least i got back to the beginning of it lol

Answer 6

are you in calculus?

Answer 7

im in calc2 at the moment, but have taught myself alot more

Answer 8

@JaredP:Sqeeze theorem is not much help for competitive mathematics (say entrance exams) as L'hopitals rule is much faster.however,you could use math.stackexchange.com for a rigorous discussion.

Answer 9

http://www.wolframalpha.com/input/?i=y%3Dsqrt%28x^2%2B1%29+and+y%3Dx

if i were to use the squeeze thrm tho id prolly go this route

Answer 10

@amistre64 Ya I am doing Calculus 1 right now via correspondance. I have been out of school for 7 years, it's taking me quite a lot of time to figure this stuff out, but it's slowly making sense. At first I thought I was going to die! ;)

Answer 11

keep at it, it gets better and funner towards the end ;)

good luck