I need to find the horizontal asymptote for this function:
y= x/((sqrt)x^2-1)
the method we're supposed to use is multiple the numerator and denominator by (1/x). I just have some problems when square roots are involved.
As long as you don't have to be super formal in your method, here's a way to do it.
$\displaystyle \sqrt{x^2-1}=\sqrt{x^2(1-\frac{1}{x^2})}=\sqrt{x^2}*\sqrt{(1-\frac{1}{x^2})}=x*\sqrt{(1-\frac{1}{x^2})}$
Now you can cancel out your x's and evaluate the new limit as x approaches infinity of $\displaystyle \frac{1}{\sqrt{1-\frac{1}{x^2}}}$, which should be trivial.