# Confused with example for textbook, Dividing sqrt help pleaseee

• Aug 10th 2010, 06:12 PM
kensington
Confused with example for textbook, Dividing sqrt help pleaseee
k so in my textbook i see this example
lim as x approaches infinity sqrt(2x^2+1)/(3x-5)
and it explains how u divide the numerator and denominator by the highest power of x in the denominator. Everything i understood until it showed:

lim x approaches infinity sqrt(2x^2+1)/(3x-5) =lim x approaches infinity sqrt(2+(1/x^2))/(3-5/x)

So the denominator i understand, simple cancelling of the x, but how did we go from sqwrt(2x^2+1) to sqrt (2+(1/x^2)) just by dividing by x? So confused
• Aug 10th 2010, 06:18 PM
skeeter
Quote:

Originally Posted by kensington
k so in my textbook i see this example
lim as x approaches infinity sqrt(2x^2+1)/(3x-5)
and it explains how u divide the numerator and denominator by the highest power of x in the denominator. Everything i understood until it showed:

lim x approaches infinity sqrt(2x^2+1)/(3x-5) =lim x approaches infinity sqrt(2+(1/x^2))/(3-5/x)

So the denominator i understand, simple cancelling of the x, but how did we go from sqwrt(2x^2+1) to sqrt (2+(1/x^2)) just by dividing by x? So confused

since x > 0, each term in the the numerator was divided by $x = \sqrt{x^2}$
• Aug 10th 2010, 06:20 PM
tonio
Quote:

Originally Posted by kensington
k so in my textbook i see this example
lim as x approaches infinity sqrt(2x^2+1)/(3x-5)
and it explains how u divide the numerator and denominator by the highest power of x in the denominator. Everything i understood until it showed:

lim x approaches infinity sqrt(2x^2+1)/(3x-5) =lim x approaches infinity sqrt(2+(1/x^2))/(3-5/x)

So the denominator i understand, simple cancelling of the x, but how did we go from sqwrt(2x^2+1) to sqrt (2+(1/x^2)) just by dividing by x? So confused

For positive x, you have $x\sqrt{a}=\sqrt{x^2a}$, so $\frac{1}{x}\,\sqrt{2x^2+1}=\sqrt{\frac{2x^2+1}{x^2 }}=\sqrt{2+\frac{1}{x^2}}$

Tonio