Posted by Sarah on Saturday, March 13, 2010 at 8:48pm.
how do you find the limit at infinity of:
lim(x>infinity) (x+2)/sqrt(64 x^2+1)
Do you first change the square root on denominator to (64x^2+1)^1/2 and then divide everything by zero.
Please help me with this, I'm confused.

Calculus  bobpursley, Saturday, March 13, 2010 at 9:33pm
Divide by zero? That is not allowed here in Texas.
expand the denominator by the binomial expansion...
(64x^2+1)^.5 = (64x^2)^.5+ Cn,k(64x^2)^.5+Cn,k(64x^2)^1.5+....
well, as x>inf, all the terms after the first go to zero,
So the denominator becomes 8x
Limx>inf (x+2)/(8x+ ...)= 1/8
One other way. Multipy numerator and denominator by 1/x
(1+2/x)/sqrt(64+1/x^2)
now, as the limit as x>large then is
1/sqrt64= 1/8