# Thread: How do u find the Horizonal asympotes of this

1. ## How do u find the Horizonal asympotes of this

y= x
sqrtx^2 -4

2. Originally Posted by johntuan
y= x
sqrtx^2 -4
Find $\displaystyle \lim_{x \to \infty}$ and $\displaystyle \lim_{x \to - \infty}$, if you get a finite result, that is a horizontal asymptote

and please type the function more clearly. Is it $\displaystyle y = \frac {x}{ \sqrt {x} - 4} \mbox { , } \frac {x}{ \sqrt {x - 4}} \mbox { or } \frac {x}{\sqrt {x}} - 4$ ?

It would make no sense to write the last one like that, but I've seen things like that before

3. sorry its the middle one

4. Originally Posted by johntuan
sorry its the middle one
If you can't use LaTex, you would type it as y = x/[sqrt(x - 4)]. So, what are the limits?

5. the limits are these $\displaystyle \lim_{x \to \infty}$ and $\displaystyle \lim_{x \to - \infty}$. The only thing that is messing me up is the sqrt sign.

6. Originally Posted by johntuan
the limits are these $\displaystyle \lim_{x \to \infty}$ and $\displaystyle \lim_{x \to - \infty}$. The only thing that is messing me up is the sqrt sign.
No, no, no. those are limit notations my friend. I want you to find:

$\displaystyle \lim_{x \to \infty} \frac {x}{\sqrt {x - 4}} =$ ?

and

$\displaystyle \lim_{x \to - \infty} \frac {x}{ \sqrt {x - 4}} =$ ?

Of course the second limit is kind of trivial, since the domain of this function is $\displaystyle (4, \infty)$. As $\displaystyle x \to 4$, the denominator goes to 0, so the function will shoot up to positive infinity, with a vertical asymptote of x = 4. So what is the first limit? This will give you the horizontal asymptote if there is one

Hint: For very large x values, the -4 will be inconsequential. So the function will behave like $\displaystyle \frac {x}{\sqrt {x}}$ for very large x's

7. whoops sorry the function is $\displaystyle \lim_{x \to \infty} \frac {x} {\sqrt {x^2 - 4}} =$ I got the VA of +2 and -2 but for the HA ok if I plug in $\displaystyle \infty$ I would get infinity over sqrt infinity^2 - 4. Which would then lead to infinity over infinity?

8. Originally Posted by johntuan
whoops sorry the function is $\displaystyle \lim_{x \to \infty} \frac {x} {\sqrt {x^2 - 4}} =$ I got the VA of +2 and -2 but for the HA ok if I plug in $\displaystyle \infty$ I would get infinity over sqrt infinity^2 - 4. Which would then lead to infinity over infinity?
Now that's a major typo! It changes the answer completely. You are right for the vertical asymptotes. Note that the function is undefined on the interval $\displaystyle [-2,2]$

Anyway, the same concept applies. For large x's (both negative and positive), the -4 is inconsequential. So,

$\displaystyle \lim_{x \to \infty} \frac {x}{\sqrt{x^2 - 4}} = \lim_{x \to \infty} \frac {x}{\sqrt{x^2}} =$ ?

and

$\displaystyle \lim_{x \to - \infty} \frac {x}{\sqrt {x^2 - 4}} = \lim_{x \to - \infty} \frac {x}{\sqrt {x^2}} =$ ?

Now what do you think?

EDIT: My bad. It turns out you had x^2 to begin with, I was the one who misread the question. Sorry

9. Originally Posted by Jhevon
Now that's a major typo! It changes the answer completely. You are right for the vertical asymptotes. Note that the function is undefined on the interval $\displaystyle [-2,2]$

Anyway, the same concept applies. For large x's (both negative and positive), the -4 is inconsequential. So,

$\displaystyle \lim_{x \to \infty} \frac {x}{\sqrt{x^2 - 4}} = \lim_{x \to \infty} \frac {x}{\sqrt{x^2}} =$ ?

and

$\displaystyle \lim_{x \to - \infty} \frac {x}{\sqrt {x^2 - 4}} = \lim_{x \to - \infty} \frac {x}{\sqrt {x^2}} =$ ?

Now what do you think?
I was hoping you would help me on this part but I'll give it a shot. Would it become infinity over sqrt infinity.

If we plug in negative infinity would it become negative infinity over sqrt negative infinity? Which would make both answers meaningless?

10. Originally Posted by johntuan
I was hoping you would help me on this part but I'll give it a shot. Would it become infinity over sqrt infinity.

If we plug in negative infinity would it become negative infinity over sqrt negative infinity? Which would make both answers meaningless?
Simplify first!

Recall that, by definition, $\displaystyle \sqrt {x^2} = |x|$ which is always nonnegative.

This should not be that surprising, since $\displaystyle x^2$ will always be nonegative, so $\displaystyle \sqrt {x^2}$ must be nonnegative as well. We can only take the positive values this function gives however, since the denominator can't be 0 either. With that in mind, let us proceed.

$\displaystyle \lim_{x \to \infty} \frac {x}{\sqrt {x^2 - 4}} = \lim_{x \to \infty} \frac {x}{\sqrt {x^2}} = \lim_{x \to \infty} \frac {x}{|x|}$

Now, from our knowledge of absolute valued functions, we know that:
$\displaystyle \frac {x}{|x|} =\left\{\begin{array}{cc}1,&\mbox{ if } x> 0\\-1, & \mbox{ if } x<0\end{array}\right.$

And it is indeterminate otherwise.

So our limit as x gets large in the positive direction becomes:

$\displaystyle \lim_{x \to \infty} 1 = 1$

Similarly, we will find that $\displaystyle \lim_{x \to - \infty} \frac {x}{\sqrt {x^2 - 4}} = \lim_{x \to - \infty} -1 = -1$

So the horizontal asymptotes are $\displaystyle y = 1$ and $\displaystyle y = -1$

EDIT: I don't know how much experience you have with absolute valued functions, if there is anything you don't get, please say so.

11. thank you very much I forgot about the absolute rules