horizontal asymptotes

**calcbeg** · February 8th 2009, 04:33 PM

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

**ThePerfectHacker** · February 8th 2009, 04:56 PM

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?

Thanks in advance
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?

**calcbeg** · February 8th 2009, 05:15 PM

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

**ThePerfectHacker** · February 8th 2009, 05:24 PM

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?

**calcbeg** · February 8th 2009, 06:05 PM

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

**ThePerfectHacker** · February 8th 2009, 06:14 PM

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

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