Find the horizontal asymptote

f(x)= [sqrt(2x^2 +1)] / [3x-5]

for x-> infinity, we have x>0

I need to divide numerator and denominator by the highest power of the denominator, which is x in this case.

lim x-> infinity [sqrt(2x^2 +1)] / [3x-5]

The solution in the book has the following step

lim x-> infinity [sqrt(2 +1/x^2)] / [3-5/x]

I do not understand how they get to this numerator, when dividing by x?

The book gives also:

For x>0 --> sqrt(x^2) = x

For x<0 --> sqrt(x^2) = |x| = -x

Can someone explain that to me. What steps are in between.

Thanks