Whats the limit of?
Lim square root of ( (x^2+ax) - square root of(x^2+bx) )
x increase without bound (+ infinity)
- Mark PLv 51 decade agoFavorite Answer
The limit is infinity.
As x approaches infinity, you can throw out the ax and bx terms because x^2 dominates.
So sqrt(x^2-sqrt(x^2)) = sqrt(x^2 - x)
Again, the x^2 dominates, so you simply have:
lim(x->infinity) of x
which is simply infinity.
Now, if you misplaced your parentheses in the problem statement, the answer is more interesting.
lim(x->inf) of sqrt(x^2+ax) is x+a/2, since (x+a/2)^2 = (x^2+ax+1)
lim(x->inf) of sqrt(x^2+bx) is x+b/2
So lim(x->inf) of the difference is (x+a/2)-(x+b/2) or (a-b)/2.
This is confirmed with calculations using very large x and small a and b.
- Phineas BoggLv 61 decade ago
I believe it is (a-b)/2.
x^2 + ax + (a/2)^2 = (x+a/2)^2 and
x^2 + bx + (b/2)^2 = (x+b/2)^2
So the limit looks like limit:
sqrt[(x+a/2)^2] - sqrt[(x+b/2)^2] =
x+a/2 - (x+b/2) = a/2 - b/2
It's not formal, but if I had to put money on this problem, I would put it on (a-b)/2
- MafiaLv 41 decade ago
if lim of equation when x approaches infinity, then limit is rad a - rad b.