# Thread: Help with infinite limits

1. ## Help with infinite limits

I need a good explanation on how to complete the following problem.

The limit as x approaches infinity...(64x^2 +x "which is under a square root"
- 8x)

Thanks very much

2. $\displaystyle \lim_{x\to\infty}\bigl(\sqrt{64x^2+x}-8x\bigr)$

There is a standard trick for dealing with expressions like this. Make it into a fraction, multiplying top and bottom by the "conjugate" expression (got by changing the – sign to a +):

$\displaystyle \sqrt{64x^2+x}-8x = \frac{\bigl(\sqrt{64x^2+x}-8x\bigr)\bigl(\sqrt{64x^2+x}+8x\bigr)} {\sqrt{64x^2+x}+8x}$

Multiply out the numerator (difference of two squares) and see if you can figure out what then happens as x→∞.