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