1. ## horizontal asymptotes

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?

this calculus beginner

2. Originally Posted by calcbeg
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?

this calculus beginner
To find the horizontal asymptotes you need to compute:
$\lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x-2}\text{ and }\lim_{x\to -\infty}\frac{\sqrt{x^2+1}}{x-2}$.

Can you compute those limits?

3. I am finding it hard to deal with the square root signs so if you can walk me through it I would appreciate it.

Thanks

4. Originally Posted by calcbeg
I am finding it hard to deal with the square root signs so if you can walk me through it I would appreciate it.

Thanks
$\frac{\sqrt{x^2+1}}{x-2} = \frac{\sqrt{x^2+1}}{x-2} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}} = \frac{\sqrt{1 + \tfrac{1}{x^2}}}{1 - \tfrac{2}{x}}$ for $x>0$.
What happens when $x\to \infty$?

If $x<0$ then we need to be careful we cannot multiply both the numerator and denominator with $\tfrac{1}{x}$ because $\tfrac{1}{x}$ and you cannot put a negative number under a square root. Rather you need to multiply the denominator by $\tfrac{1}{-x}$. Can you do the second limit?

5. 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

6. Originally Posted by calcbeg
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 $x\to \infty$ means $x$ gets almost as large as my ego.
Thus, if $x$ is getting large it would mean $\tfrac{1}{x}$ goes to zero, as well as $\tfrac{1}{x^2}\to 0$.

Now, $\lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x-2} = \lim_{x\to \infty}\frac{\sqrt{x^2+1}}{x-2} \cdot \frac{\tfrac{1}{x}}{\tfrac{1}{x}} = \lim_{x\to \infty} \frac{\sqrt{1 + \tfrac{1}{x^2}}}{1 - \tfrac{2}{x}} = \frac{\sqrt{1+0}}{1-0} = 1$