# Find the horizontal asymptote

• Oct 26th 2009, 01:26 PM
DBA
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
• Oct 26th 2009, 03:12 PM
ANDS!
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"?
• Oct 26th 2009, 03:13 PM
Jameson
Quote:

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.