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.