What is the limit of g(x) when....?

Question

x-> infinity and g(x) = (sqrt(x^2 + ax)) - (sqrt(x^2 +bx)) ???
explanation please the answer is (a-b)/2

Answer 1

multiply top and bottom by the conjugate.
(sqrt(x^2 + ax)) - (sqrt(x^2 +bx)) (sqrt(x^2 + ax)) + (sqrt(x^2 +bx))/ (sqrt(x^2 + ax)) +(sqrt(x^2 +bx))

you get: (x^2 + ax - x^2 - bx) / (sqrt(x^2 + ax)) +(sqrt(x^2 +bx))

= (ax - bx)/ (sqrt(x^2 + ax)) +(sqrt(x^2 +bx))

= (a - b) / (sqrt(x^2 + ax)) +(sqrt(x^2 +bx)) / x

= (a - b) / (sqrt(x^2 + ax)/x^2) +(sqrt(x^2 +bx)/x^2)

= (a - b) / [sqrt ( 1 + a/x^2) + sqrt ( 1 + b/x^2)

The denominator approaches 2 as x approaches oo, since a/x^2 and b/x^2 go to 0 so the limit is (a - b)/2

Answer 2

Girl Nobody knows how to do that stuff.
Go to a homework web site or use the CD that came with the book.
Do they have a homework hotline in your town.
Call your teacher I bet she is in the white pages.