Posted by John on Saturday, November 22, 2008 at 11:54pm.
Limit as x goes to negative infinity of sqrt(x^2+3x)+x?

Calculus  drwls, Sunday, November 23, 2008 at 12:16am
It depends upon whether you take the positive or negative square root. If you take the positive root and x> infinity, the first term approaches = x = x (for negative x). Adding x to that gives you zero in the limit.
Check, let x = 1000
sqrt(10^6 3000)= 999.5
Subtract 1000 and you get 0.5
Let x = 10,000
sqrt(10^8 3^10^4) = 999.85
Suntract 10,000 and you get 0.15 
Calculus  John, Sunday, November 23, 2008 at 12:40am
I still don't get it. The answer is 1.5, how do I get that w/o a calculator?

Calculus  drwls, Sunday, November 23, 2008 at 6:29am
You are correct, both my derivation and my numerical examples were wrong. You can get 1.5 by considering a series expansion of sqrt(1+y) = 1 + y/2 + ...
sqrt(x^2 + 3x) = sqrt[x^2*(1 + (3/x)]
= x * [1 + (3/2)x + higher order terms in 3/x] = x + 3/2 + ...
As x approaches infinity, with positive square roots only taken, the first term sqrt(x^2) approaches x and the second term remains 3/2
Adding x to that is like subtracting
x, leaving you with 3/2
Fox x = 1000, f(x) = 1.5011
for x = 10000, f(x) = 1.5001
I apologize for my sloppy math.