y= x
sqrtx^2 -4
No, no, no. those are limit notations my friend. I want you to find:
?
and
?
Of course the second limit is kind of trivial, since the domain of this function is . As , 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 for very large x's
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
Anyway, the same concept applies. For large x's (both negative and positive), the -4 is inconsequential. So,
?
and
?
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, which is always nonnegative.
This should not be that surprising, since will always be nonegative, so 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.
Now, from our knowledge of absolute valued functions, we know that:
And it is indeterminate otherwise.
So our limit as x gets large in the positive direction becomes:
Similarly, we will find that
So the horizontal asymptotes are and
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.