Calculus
posted by Sarah .
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.

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
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