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.