# Thread: Find the horizontal asymptote

1. ## 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

2. As you noted, you want to divide by the highest power in the denominator which is x. However, "x" isn't the only way to represent "x". Is not $\sqrt{x^{2}}$ (for x greater than 0) also a means of denoting "x"?

3. Originally Posted by DBA
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?
You are missing an important term, but this is almost correct.

What they did is factor out an x^2 from x^2+1, which becomes x^2(1+[1\x^2]). Now you can apply the square root to the x^2 term and demonstrate that the powers of the numerator and denominator are equal. Also notice that 1/x^2 tends to 0 as x goes to infinity.