Ordinarily, I wouldn't post a math problem to RMR, but I'm hoping that
>Duane Wilkey is scanning this. I just got a new computer and don't have
>his e-mail address. So, if anyone can e-mail me Duane's e-mail
>address...or solve the problem, I'd be greatful.
>
>here goes:
>Sqrt(X2 + X +1) - sqrt(X2-X) Limit as X goes to infinity
>
Do you mean sqrt(x^2 + x + 1) - sqrt(x^2 - x) ? If so, I think it's 1.
Couldn't show you the proof, but trying it with x=10^6 on my lil' ol' HP results is 1, just as Darren suggests.
MACSYMA is a free download these days...
BillW
Thank you, Dr. Ritz. I was trying to remember what your doctorate was in when I relayed your solution to my son. Was it math, by chance?
Mark Simpson NAR 71503 Level II God Bless our peacekeepers
Mark Simpson wrote: > John Ritz wrote >> If you mean Sqrt(X^2+X+1)-Sqrt(X^2-X), yes then it goes to one : >>
John has a clever solution, and it looks right. But it doesn't demonstrate L'Hospital's Rule, which is
f(x) f'(x) lim ---- = lim ----- g(x) g'(x)
if both lim f(x) and lim g(x) are 0 or infinity.
Mark, your son was on the right track:
So make f(x) = (2x+1) and g(x) = (sqrt(x^2 + x + 1) + sqrt(x^2 - x)); then f'(x) is 2, and g'(x) is:
(2x + 1)/2*sqrt(x^2 + x + 1) + (2x - 1)/2*sqrt(x^2 - x)
As x -> infinity, g'(x) is approximately
(2x)/2*sqrt(x^2) + (2x)/2*sqrt(x^2) = 2
(see John's email for why) and lim f'(x)/g'(x) is 1.
Mark,
There's a problem here; can you provide the entire original equations?
L'Hospital's Rule is normally used to find a limit when a quotient of functions assumes an indeterminate form, such as 0/0, as a variable is allowed to go to a limit, like infinity in this case.
Are these terms the result of a differentiation using L'Hospital's Rule, or are you just trying to find the limit of this expression?
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