Hi again
f(x) = (√(x^2+1)/(x-2))
I think the vertical asymptote is 2 but what is the horizontal asymptote?
How do you figure this out?
Thanks in advance
this calculus beginner
$\displaystyle \frac{\sqrt{x^2+1}}{x-2} = \frac{\sqrt{x^2+1}}{x-2} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}} = \frac{\sqrt{1 + \tfrac{1}{x^2}}}{1 - \tfrac{2}{x}}$ for $\displaystyle x>0$.
What happens when $\displaystyle x\to \infty$?
If $\displaystyle x<0$ then we need to be careful we cannot multiply both the numerator and denominator with $\displaystyle \tfrac{1}{x}$ because $\displaystyle \tfrac{1}{x}$ and you cannot put a negative number under a square root. Rather you need to multiply the denominator by $\displaystyle \tfrac{1}{-x}$. Can you do the second limit?
The examples I have found seem to show getting rid of the x's on the top of the formula but you haven't done that with the last one. I am expecting to come up with a number and I don't know what that is on the first one let alone the second one. Do I substitute 0 for x at some point?
Thanks
The meaning of $\displaystyle x\to \infty$ means $\displaystyle x$ gets almost as large as my ego.
Thus, if $\displaystyle x$ is getting large it would mean $\displaystyle \tfrac{1}{x}$ goes to zero, as well as $\displaystyle \tfrac{1}{x^2}\to 0$.
Now, $\displaystyle \lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x-2} = \lim_{x\to \infty}\frac{\sqrt{x^2+1}}{x-2} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}} = \lim_{x\to \infty} \frac{\sqrt{1 + \tfrac{1}{x^2}}}{1 - \tfrac{2}{x}} = \frac{\sqrt{1+0}}{1-0} = 1$