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
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
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
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.