it's calculus time!!! YES!!!

[Alex]

this week was way better than the last and i only have 1 doubt that i'd like clarified, if possible:

to find: lim (x^2 + 2)^0.5 / (2x^2 + 1)^0.5
x->infinity


edit: sorry forgot the main part of the question, so somehow this evens out to be infinity so we use L'Hopital's Rule:

so the answer is:

f ' ((x^2 + 2)^0.5) / g ' ((2x^2 + 1)^0.5)

im absolutely at a loss about how to tackle this one, as it just gets way out of hand way too quick with the ^-.5 etc...

anybody know this?

thx!

 

[tontod]

Its been a few years since I last took Calculus, but my guess the answer is 1 (infinity/infinity). Someone correct me if I'm wrong.

 

[Alex]

yeah thats the answer, but how did you find it?

and the next step is differentiating it...

 

[XZeroII]

My oppinion of how it would be done goes like this:
X goes to infinity. Infinity squared is infinity. The +2 or +1 doesn't matter at all. The square root of infinity doesn't matter because square root of infinity is infinity. Thus it would be infinity over infinity. Use the quotient rule to differentiate.
(f'(x)g(x) - f(x)g'(x)) / g(x)^2
f(x) = numerator
g(x) = denominator


EDIT: Fixed my quotient rule

 

[tontod]

Woohoo, didnt think I remembered any of this stuff. What you do is for (x^2 + 2), as x goes to infinity, you can discount 2 and square root of infinity can be considered to be infinity. So the numerator is infinity, same way the denominator is infinity. Differentiating it shouldnt be too difficult I would imagine, I'm having trouble remembering how to do it exactly.

 

[BigPoppa]

The answer is 1/sqrt(2). You obtain this by dividing top and bottom by the highest x power. x^2/x^2 = sqrt(1) or 1. 2x^2/x^2 = sqrt(2). Hope that helps.

 

[Alex]

hmmmmm

 

[FeathersMcGraw]

Sounds like you're making a common mistake in trying to apply L'Hopital's rule, which is that you're not differentiating a quotient, you're taking the limit of a quotient of derivatives. Separately differentiate f(x) = (x^2 + 2)^(1/2) and g(x) = (2x^2 + 1)^(1/2), then find the limit of f'(x)/g'(x) as x -> oo.

 

[Alex]

Originally posted by: FeathersMcGraw
Sounds like you're making a common mistake in trying to apply L'Hopital's rule, which is that you're not differentiating a quotient, you're taking the limit of a quotient of derivatives. Separately differentiate f(x) = (x^2 + 2)^(1/2) and g(x) = (2x^2 + 1)^(1/2), then find the limit of f'(x)/g'(x) as x -> oo.

yeah thats exactly what i did!

 

[fumbduck]

i just finished calculus AB. And i know how to do this problem and could easily do it in a few minutes. However, I am lazy. Sorry.

 

[josphII]

Originally posted by: tontod
Its been a few years since I last took Calculus, but my guess the answer is 1 (infinity/infinity). Someone correct me if I'm wrong.

FYI - inf/inf does not equal 1

 

[FeathersMcGraw]

Originally posted by: josphII

FYI - inf/inf does not equal 1

In general, that's certainly true, but lim(x->oo) x/x = oo/oo certainly does equal 1.

 

[josphII]

Originally posted by: FeathersMcGraw

Originally posted by: josphII

FYI - inf/inf does not equal 1

In general, that's certainly true, but lim(x->oo) x/x = oo/oo certainly does equal 1.

infinity/infinity always equals undefined. your example is flawed because x/x reduces to 1 w/out any substitution. you can prove that the lim as x->inf of x/x equals 1 by applying l'hopitals rule, not by simple substitution