# Calculus

posted by .

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.

• Calculus -

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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