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?
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.