Domains and Asymptotes

Deter...

Question

Domains and Asymptotes

Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.

y = (√(x² + 4)) / x

Accepted Answer

Welcome back, everyone. Determine the domain and asymptotes of the given function. Y equals square root of X2 + 1 divided by X. Let's begin with the domain. We can see that we're given a rational function, and we have to recall that division by 0 is undefined for rational functions in a form of P X divided by Q of X, right? So whenever we have this form, we want to ensure that our denominator is never equal to 0. So our denominator is X, meaning X cannot be equal to 0, and this gives us our domain. So for the domain part, we can say that x belongs to the interval of all real numbers except from 0, which can be written as negative infinity up to 0, union 0 up to infinity. So we have our first answer and we can label it. Now let's identify the asymptotes. We can begin with with the vertical asymptote, and we have to recall that the vertical asymptote makes our denominator equal to 0, so we're going to set Q of X equal to 0, and we want to ensure that our numerator is not equal to 0 at the same value of X. Otherwise, we might have a potential hole, right? So, for the vertical asymptote, if we want to set our denominator equal to 0, this means that X is equal to 0. And square root of x2 + 1 should not be equal to 0. So now if we substitute X equals 0 into the second expression, we get square root of 0 + 1, which is 1, and that's not equal to 0, meaning this squarifies that x equals 0 is the vertical asymptote. That's our second answer to this problem, and now let's consider the. Horizontal asymptotes. For the horizontal asymptotes we want to identify two limits. Limit as x approaches positive infinity. Let's begin with a positive one of the given function. So we have square root of X2 + 1 divided by X. If X approaches infinity, we simply get an indeterminate form of infinity divided by infinity. And so so for this limit what we can do is first we'll factor out X2d under the radical. This gives us limit as x approaches infinity of square root of X2, then print C 1 + 1 divided by X2. In the denominator we have X. And now we can take square root of X2, which gives us the absolute value of X. So we get the absolute value of X multiplied by square root of 1 + 1 divided by X 2 divided by X. Now because X is positive, since it approaches positive infinity, we can drop the absolute value and write our limit as limit as x approaches infinity of X divided by X, so these terms cancel each other out, and we're only left with the radical term square root of 1. Plus 1 divided by X squared. And now the fractional term approach is 0 because we're dividing 1 by an infinitely large number, and square root of 1 gives us a limit of 1. And now our second horizontal asymptote will be found by identifying the limit as x approaches negative infinity. So we can use our previous form and it because X approaches negative infinity. The absolute value of X is going to be negative X based on the definition of the piece wise function. So we take negative X multiplied by a square root of 1 + 1 divided by x2d. All of that divided by X. We can cancel out the x terms and we can see that the main difference here is that we're taking minus square root of 1, which gives us -1. So there are two horizontal asymptotes Y equals 1. From the first limit and Y equals -1 from the second limit, those are the horizontal asymptotes. Well then, we have our third answer, and we can finish the problem. Thank you for watching.

Domains and Asymptotes

Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.

y = (√(x² + 4)) / x

Key Concepts

Domain of a Function

Limits and Asymptotes

Square Root Function

Watch next