Limit

lashko · 2009-08-13 10:35:35

\lim_{x\to infinity} \, \sqrt{x^2+x}-x

Last edited by lashko (2009-08-13 10:43:41)

mathsyperson · 2009-08-13 11:07:28

Can't think how to do this properly, but here's my "casual logic":

Trying x=10^6 on a calculator supports my guess.

Ricky · 2009-08-13 13:34:30

L'Hopital is the way to go. Do some algebraic manipulation till you get it into this form:

And note that the limit of the right factor is zero. Now L'Hopital, and then a little bit more algebraic manipulation and you're done.

The limit actually turns out to be about

Riad Zaidan · 2009-08-13 16:09:19

Hi lashko
I think that you can find the limit as follows:

lim( (x^2+x)^ ½ - x)multiplying both num. and den. by ( (x^2+x)^ ½+x) we get
x ⇒ ∞

( (x^2+x)^ ½ - x).[( (x^2+x)^ ½+x)
= lim _______________________________
x ⇒ ∞ ( (x^2+x)^ ½+x)

x^2+x-x^2
= lim ___________
x ⇒ ∞ ( (x^2+x)^ ½+x)

x
= lim ___________ by taking x^2 out of the root
x ⇒ ∞ ( (x^2+x)^ ½+x)

x
= lim ________________ taking x as acommon factor from the denominator
x ⇒ ∞ x(1+(1/x))^ ½+1)

1
= lim ______________ = 1/2 as x ⇒ ∞ as requiered
x ⇒ ∞ (1+(1/x))^½ +1)

Best Regards
Riad Zaidan

lashko · 2009-08-14 05:59:21

Thanks guys you are right thats 1/2

Last edited by lashko (2009-08-14 06:13:46)

lashko · 2009-08-14 06:33:39

No L'Hopital does not apply. Ive thought about it

Ricky · 2009-08-14 12:57:13

lashko wrote:

No L'Hopital does not apply. Ive thought about it

Huh? The most you can say is "I don't see how to apply it".

Math Is Fun Forum