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Calculus1 9 Online
OpenStudy (anonymous):

lim x -> infinity (sqrt(x^2+5x) -x) L'hopital Explain Please

hartnn (hartnn):

put x=1/y then y->0 then you can apply L'Hopitals.

OpenStudy (anonymous):

@hartnn isnt it somewhat the same question we answered yesterday

hartnn (hartnn):

maybe, i don't remember.....

terenzreignz (terenzreignz):

\[\lim_{x \rightarrow \infty}\sqrt{x^{2}+5x}-x\] An indeterminate form ∞-∞ There might be a "more proper" way to do this, but here's my hunch: Express it as \[\lim_{x \rightarrow \infty}\sqrt{x^{2}\left( 1-\frac{5}{x} \right)}-x\]and then \[\lim_{x \rightarrow \infty}x \sqrt{ 1-\frac{5}{x}}-x\]NOTE that it's normally |x| outside the radical and not just x, but seeing as x is going to positive infinity anyway...\[\lim_{x \rightarrow \infty}x \left( \sqrt{ 1-\frac{5}{x}}-1 \right)\]So now, it's an indeterminate of the form 0· ∞ From which the use of L'Hôpital's rule can be more clearly seen :D

OpenStudy (anonymous):

multiply by the conjugate, then it will be easier

OpenStudy (anonymous):

you get \[\frac{5x}{\sqrt{x^2+5x}+x}\] now you don't need l'hopital, your eyeballs will do

OpenStudy (anonymous):

\(\frac{5x}{2x}=\frac{5}{2}\)

OpenStudy (anonymous):

i think this one is just 5...hmm: when you multiply by the conjugate wouldn't it be √(x^2 +5x) multiply by conjugate (√(x^2+5x) +x) both numerator and denominator x^2-√(x^2+5x) / x+ √(x^2 +5x) then you simplify and you get 5x/ (x+√(x^2+5x) Find the derivative 5/(1+1/2 (x^2+5x)^-1/2 (2x+5) 5/1+[(2x+5)/2√((x^2)+5x)

OpenStudy (anonymous):

also, (2x+5)/2√((x^2)+5x) all of this is equal to 1 so, based on that ...lol i think i just proved myself wrong. I multiplied the 1's in the denominator instead of adding Yes, its 5/2 5/(1+1) =5/2

OpenStudy (anonymous):

what happened to the -x at the end of the original equation karisos

OpenStudy (anonymous):

I placed it in front of it but forgot to carry the negative sign so just leave it in the back. Bringing it upfront will just make things worse. I think the sign might make the answer negative. I'm pretty sure its 5/2 but i'm not completely sure if its -5/2. Since I forgot to carry the negative out it probably is -5/2. :/

OpenStudy (anonymous):

the negative would make the 1on the denominator negative making both ones cancel out. So in the end, I'm back to the beginning now it looks like its just 5. I'm probably confusing u more. Sorry man

OpenStudy (anonymous):

Here, I went on google and found this. Same question apparently...the answer there was 5/2 http://answers.yahoo.com/question/index?qid=20081102155643AA6WX8Z

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