Video: Limits and Asymptotic Behavior

In this video, we will learn how to use limits to understand asymptotic behavior of functions.

11:33

Video Transcript

In this video, we’re going to learn how to use limits to help us understand the asymptotic behaviour of functions. We’ll recall what it actually means for a function to have horizontal and vertical asymptotes and use the laws of limits to help us find the location of any asymptotes.

We begin by recalling the definition of a horizontal or vertical asymptote. They are, respectively, horizontal and vertical lines such that the distance between the curve and the line approaches zero as either the 𝑥- or 𝑦-coordinate approaches ∞. The graph of 𝑦 equals one over 𝑥 squared, for example, has a horizontal asymptote given by the equation 𝑦 equals zero and a vertical asymptote given by the equation 𝑥 equals zero.

It’s important to realise that a function can have absolutely zero horizontal or vertical asymptotes. For example, 𝑦 equals sin of 𝑥 and indeed 𝑦 equals cos of 𝑥 have no asymptotes at any point. Similarly, the graphs of polynomials are smooth and continuous, with no asymptotes of any kind. What we’re looking to do is formalise this definition using limits. If we let 𝑓 be defined on some open interval 𝑎 to ∞, then if the limit as 𝑥 approaches ∞ of the function is equal to some constant value 𝐿, this means that the value of the function can be made arbitrarily close to 𝐿 by requiring 𝑥 to be sufficiently large. Visually, that could look a little something like this.

Alternatively, we can say that, for a function to find on some open interval negative ∞ to 𝑎, then if the limit as 𝑥 approaches negative ∞ of the function is equal to some constant 𝐿. Then the values of the function can be made arbitrarily close to 𝐿, this time requiring 𝑥 to be sufficiently large negative. And once again, we have a diagram representing that scenario. Notice that each time, the graph approaches the horizontal line 𝑦 equals 𝐿 as either 𝑥 gets sufficiently large or sufficiently large negative. And that helps us with the formal definition for a horizontal asymptote. We say that the line given by the equation 𝑦 equals 𝐿 is a horizontal asymptote of the curve 𝑦 equals 𝑓 of 𝑥 if either the limit as 𝑥 approaches ∞ of 𝑓 of 𝑥 equals 𝐿 or the limit as 𝑥 approaches negative ∞ of 𝑓 of 𝑥 equals 𝐿.

And what about our vertical asymptotes? This time, we say that the line 𝑥 equals 𝑎 is a vertical asymptote of 𝑦 equals 𝑓 of 𝑥 if at least one of the following is true. We can look at one-sided limits. If the limit as 𝑥 approaches 𝑎 from the left of 𝑓 of 𝑥 is either positive or negative ∞. If the limit as 𝑥 approaches 𝑎 from the right of 𝑓 of 𝑥 is either positive or negative ∞. Or simply, if the limit as 𝑥 approaches 𝑎 itself of 𝑓 of 𝑥 is either positive or negative ∞. And this is for a function to find on some open interval that contains 𝑎 except possibly at 𝑥 equals 𝑎. In particular, with rational functions, those written as quotients of two functions, vertical asymptotes are vertical lines which corresponds to these zeros of the denominator of our rational function. We’re now going to have a look at how to apply these definitions.

Find the horizontal and vertical asymptotes of 𝑓 of 𝑥 equals two 𝑥 cubed plus two over 𝑥 cubed minus one.

Let’s begin by recalling the definition for a horizontal and vertical asymptote. We say that the line 𝑦 equals 𝐿 is called a horizontal asymptote of the curve 𝑦 equals 𝑓 of 𝑥 if either the limit as 𝑥 approaches ∞ of 𝑓 of 𝑥 is equal to 𝐿 or the limit as 𝑥 approaches negative ∞ of 𝑓 of 𝑥 is equal to 𝐿. And we say that the vertical line 𝑥 equals 𝑎 is called a vertical asymptote of the curve 𝑦 equals 𝑓 of 𝑥 if at least one of the statements is true. More generally, with rational functions, those written as the quotients of two functions, vertical asymptotes are the vertical lines which correspond to the zeros of the denominator of the function.

So in this question, 𝑓 of 𝑥 is equal to two 𝑥 cubed plus two over 𝑥 cubed minus one. So we’ll begin by finding the limit as 𝑥 approaches ∞ of two 𝑥 cubed plus two over 𝑥 cubed minus one. We usually begin by checking whether we can do this with direct substitution. However, if we were to use direct substitution, we’d end up with ∞ over ∞, which we know to be an indeterminate form. So instead, we’re going to look to manipulate our expression somewhat. We’re going to divide both the numerator and the denominator of our quotient by the highest power of 𝑥 in the denominator. That’s 𝑥 cubed. That gives us two 𝑥 cubed over 𝑥 cubed plus two over 𝑥 cubed all over 𝑥 cubed over 𝑥 cubed minus one over 𝑥 cubed.

Well, of course, two 𝑥 cubed over two 𝑥 cubed is two. And 𝑥 cubed over 𝑥 cubed is one. And now, we see that as we let 𝑥 approach ∞, two over 𝑥 cubed and one over 𝑥 cubed each approach zero. And so the limit as 𝑥 approaches ∞ of our function is two over one, which is simply two. And in fact, it follows that as 𝑥 approaches negative ∞, we get the same result. So the line with equation 𝑦 equals two is a horizontal asymptote of our function.

We now move on to finding the vertical asymptote. We’re looking to find the value of 𝑥 such that the limit as 𝑥 approaches this value of the function is either positive or negative ∞. And of course, the way we achieve this with rational functions is we look for the zeros of the denominator. Well, the denominator of our function is 𝑥 cubed minus one. So we need to find where 𝑥 cubed minus one is equal to zero. Let’s add one to both sides of our equation and such that 𝑥 cubed is equal to one. And then, we’ll take the cube root of both sides. And we find that the real solution to the equation 𝑥 cubed minus one equals zero is 𝑥 equals one. And we found the vertical asymptote of our function. It’s 𝑥 equals one.

In our next example, we’ll look at a function that requires a little more manipulation.

Find the horizontal and vertical asymptotes of 𝑓 of 𝑥 equals one over 𝑥 minus one minus two.

We begin by recalling the definition of a horizontal and vertical asymptote. We say that the line 𝑦 equals 𝐿 is a horizontal asymptote of a function 𝑦 equals 𝑓 of 𝑥 if the limit as 𝑥 approaches either negative or positive ∞ of 𝑓 of 𝑥 is equal to 𝐿. Similarly, 𝑥 equals 𝑎 is a vertical asymptote of a function 𝑦 equals 𝑓 of 𝑥 if any of the six statements are true. That is, the limit as 𝑥 approaches 𝑎 from the left of the function is either positive or negative ∞. The limit as 𝑥 approaches 𝑎 from the right of the function is either positive or negative ∞. Or the limit as 𝑥 approaches 𝑎 of the function is positive or negative ∞.

Now, our function 𝑓 of 𝑥 is equal to one over 𝑥 minus one minus two. So we’ll begin by evaluating the limit as 𝑥 approaches ∞ of one over 𝑥 minus one minus two. And in fact, very little manipulation is required here. As 𝑥 gets sufficiently large, one over 𝑥 minus one approaches zero. And so the limit as 𝑥 approaches ∞ of one over 𝑥 minus one minus two is zero minus two, which is simply negative two. And in fact, as 𝑥 approaches negative ∞, one over 𝑥 minus one still approaches zero. So the limit as 𝑥 approaches negative ∞ of 𝑓 of 𝑥 is equal to negative two. And we find the equation 𝑦 equals negative two is a horizontal asymptote of our function.

Next, we’ll consider the vertical asymptotes. Less formally, we say that, with rational functions, that’s those written as the quotient of two functions, vertical asymptotes are the lines which correspond to the zeros of the denominator of the function. So in fact, we’re going to begin by subtracting two from one over 𝑥 minus one by creating a common denominator. We multiply two over one by 𝑥 minus one over 𝑥 minus one. And then, we can subtract the numerator. So we have one minus two times 𝑥 minus one over 𝑥 minus one. We simplify. And we find that 𝑓 of 𝑥 can be written as three minus two 𝑥 over 𝑥 minus one.

We’re interested in the values of 𝑥 so that our denominator 𝑥 minus one is equal to zero. Well, if we solve this equation for 𝑥 by adding one to both sides, we find that 𝑥 is equal to one. And so we found the horizontal and vertical asymptotes of 𝑓 of 𝑥 equals one over 𝑥 minus one minus two. The horizontal asymptote is 𝑦 equals negative two. And the vertical asymptote is 𝑥 equals one.

We’ll now consider how we might find asymptotes for a nonrational function.

Find the horizontal and vertical asymptotes of the function 𝑓 of 𝑥 equals three 𝑥 minus sin 𝑥.

We recall the horizontal line 𝑦 equals 𝐿 is an asymptote to the function 𝑦 equals 𝑓 of 𝑥 if the limit as 𝑥 approaches either positive or negative ∞ of 𝑓 of 𝑥 is equal to 𝐿. Then, for a vertical asymptote, we say 𝑥 equals 𝑎 is an asymptote if the limit as 𝑥 approaches 𝑎 from the left of 𝑓 of 𝑥 is either positive or negative ∞. If the limit as 𝑥 approaches 𝑎 from the right of 𝑓 of 𝑥 is either positive or negative ∞. Or if the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 is either positive or negative ∞.

Now, in this question, 𝑓 of 𝑥 is given by three 𝑥 minus sin 𝑥. So let’s begin by finding the limit as 𝑥 approaches ∞ of three 𝑥 minus sin 𝑥. Then, we recall that the limit of the sum or difference of two functions is equal to the sum or difference of the limits of those respective functions. And we can write this as the limit as 𝑥 approaches ∞ of three 𝑥 minus the limit as 𝑥 approaches ∞ of sin 𝑥. As 𝑥 approaches ∞, three 𝑥 itself also approaches ∞ whereas sin 𝑥 can only take values in the closed interval from negative one to one. And so the limit as 𝑥 approaches ∞ of three 𝑥 minus sin 𝑥 is ∞. This isn’t a constant value as required of 𝐿. So there’s no horizontal asymptote here.

We should check the limit as 𝑥 approaches negative ∞. Once again, we split it up into the limit of three 𝑥 minus the limit of sin 𝑥. As 𝑥 approaches negative ∞, three 𝑥 itself also approaches negative ∞. Once again though, sin 𝑥 oscillates between negative one and one. This has very little impact on a number as large as negative ∞. So the limit as 𝑥 approaches negative ∞ of three 𝑥 minus sin 𝑥 is negative ∞. And we can say that there are no horizontal asymptotes.

Now, we’ll consider the vertical asymptotes. We’ll split our limit up. And we’ll look for the limit as 𝑥 approaches 𝑎 of three 𝑥 minus the limit as 𝑥 approaches 𝑎 of sin 𝑥. We want to find a situation where this might be equal to either positive or negative ∞. Well, we saw that the only way for the limit of three 𝑥 to be ∞ is if 𝑥 itself approaches ∞. Similarly, for three 𝑥 to approach negative ∞, 𝑥 itself must approach negative ∞ whereas the range of sin 𝑥 is the closed interval from negative one to one. So the only way for the limit of our function to approach positive or negative ∞ is if 𝑥 itself approaches positive or negative ∞. So actually, there are no vertical asymptotes either. And so the function 𝑓 of 𝑥 equals three 𝑥 minus sin 𝑥 has no horizontal or vertical asymptotes.

By considering the three examples we’ve looked at so far in this video, you might be inclined to believe that any function that is not a rational function will not have horizontal or vertical asymptotes. In fact, this is not true. And we’re able to quote a couple of these. We know that the limit as 𝑥 approaches zero from the right of the function the natural log of 𝑥 is negative ∞. And so 𝑥 equals zero is a vertical asymptote for the function 𝑦 equals the natural log of 𝑥. And in fact, the same is true for 𝑦 equals log base 𝑏 of 𝑥 provided that the value of 𝑏 is greater than one.

In this video, we’ve learned that the line 𝑦 equals 𝐿 is called a horizontal asymptote of the curve 𝑦 equals 𝑓 of 𝑥 if either the limit as 𝑥 approaches ∞ of 𝑓 of 𝑥 is equal to 𝐿 or the limit as 𝑥 approaches negative ∞ of 𝑓 of 𝑥 is equal to 𝐿. Similarly, we found that the vertical line 𝑥 equals 𝑎 is a vertical asymptote of the curve 𝑦 equals 𝑓 of 𝑥 if any of the six statements shown are true. We saw that, less formally, with rational functions, vertical asymptotes correspond to the zeros of the denominator of that function.

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