Video: Limits at Infinity and Unbounded Limits

In this video, we will learn how to evaluate limits of a function when π‘₯ tends to infinity and explore unbounded limits that tend to infinity as π‘₯ approaches a certain value.

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Video Transcript

In this video, we’re going to learn about limits at infinity and unbounded limits. We’ve learnt previously about the meaning of the limit of 𝑓 of π‘₯ as π‘₯ approaches some real number 𝑐. If the value of this limit is 𝐿, this means that if we choose π‘₯ close enough to 𝑐, then we can make 𝑓 of π‘₯ as close to 𝐿 as we’d like. We can get the value of 𝑓 of π‘₯ arbitrarily close to 𝐿. In this video, we’re going to interpret limits of the form the limit of 𝑓 of π‘₯ as π‘₯ approaches infinity. What does it mean for such limits to have the value 𝐿?

We can try just replacing 𝑐 in the above definition by infinity. And so, we interpret this to mean that if we choose π‘₯ close enough to infinity, then we can make 𝑓 of π‘₯ as close to 𝐿 as we’d like. But what does it mean for π‘₯ be close enough to infinity when infinity is infinitely far away from any value of π‘₯ that we could choose? It turns out that instead of saying that π‘₯ is close enough to infinity, we should say that π‘₯ is large enough. So, the limit of 𝑓 of π‘₯ as π‘₯ approaches infinity equals 𝐿 means that if we choose π‘₯ large enough, then we can make 𝑓 of π‘₯ as close to 𝐿 as we’d like.

Looking at the graph of the reciprocal function, we can see that if we choose π‘₯ large enough, then we can make the reciprocal function, one over π‘₯, as close to zero as we’d like. And so, we say that the limit of one over π‘₯, as π‘₯ approaches infinity, is zero. The value of this limit zero is the value that the function one over π‘₯ gets closer and closer to as π‘₯ increases without bound.

We can also think about the limit of 𝑓 of π‘₯ as π‘₯ approaches negative infinity. This limit being 𝐿 means that if we choose π‘₯ large and negative enough, in other words π‘₯ is negative but large enough in magnitude, then we can make 𝑓 of π‘₯ as close to 𝐿 as we’d like. As with the limit as π‘₯ approaches positive infinity, we can think about this value 𝐿 in a different way. This value 𝐿 is the value that 𝑓 of π‘₯ gets closer and closer to as π‘₯ decreases without bound.

So, what is the limit of one over π‘₯ as π‘₯ approaches negative infinity? Well, as π‘₯ decreases without bounds, one over π‘₯ gets closer and closer to zero. So, the value of this limit, again, is zero. These two limits are very useful limits to know. It turns out that the limit laws we learnt for finite limits, suitably interpreted, work just as well for infinite limits. Using these laws of limits along with the limits of the reciprocal function as π‘₯ approaches infinity and negative infinity that we’ve just found, we can find the value of many other limits. Let’s see an example.

Find the limit of negative four over π‘₯ squared plus five over π‘₯ plus eight as π‘₯ approaches infinity.

We have a limit as π‘₯ approaches infinity here, but all the normal rules of limits still apply. For example, the limit of a sum of functions is equal to the sum of the limits. And so, we can split our limit up into three. It’s equal to the limit of negative four over π‘₯ squared as π‘₯ approaches infinity plus the limit of five over π‘₯ as π‘₯ approaches infinity plus the limit of eight as π‘₯ approaches infinity.

What can we say about this limit? Well, we know that the limit of a constant 𝐾, as π‘₯ approaches some number 𝑐, is just 𝐾. And as for the previous limit law, this holds true, even if 𝑐 isn’t a real number but is infinity or negative infinity. The value of this last limit is just eight.

What about the other two limits? We can use the fact that the limit of a constant multiple of a function is that constant multiple of the limit of the function. The first limit is, therefore, negative four times the limit of one over π‘₯ squared as π‘₯ approaches infinity. And the second is five times the limit of one over π‘₯ as π‘₯ approaches infinity. And finally, we add the eight.

Now, the limit of the reciprocal function one over π‘₯, as π‘₯ approaches infinity, is something we should know. Its value is zero. But how about the limit of one over π‘₯ squared as π‘₯ approaches infinity? Well, we can use the fact that the limit of a power of a function is that power of the limit of the function. This limit is the limit of the reciprocal function one over π‘₯ squared, as one over π‘₯ squared equals one over π‘₯ all squared. And by our limit law, this is the limit of one over π‘₯ as π‘₯ approaches infinity all squared. This limit is known to be zero. And so, our limit, the limit of one over π‘₯ squared as π‘₯ approaches infinity, is also zero.

We can generalize in that you get another limit law that the limit of one over π‘₯ to the power of 𝑛, as π‘₯ approaches infinity, is zero, at least if 𝑛 is greater than zero. Our original limit is, therefore, negative four times zero plus five times zero plus eight, which is, of course, just eight. Let’s see another example.

Find the limit of negative two π‘₯ to the negative four plus eight π‘₯ to the negative three minus π‘₯ to the negative two plus nine π‘₯ to the negative one minus four all over two π‘₯ to the negative four minus six π‘₯ to the negative three plus seven π‘₯ to the negative two plus six π‘₯ to the negative one plus three, as π‘₯ approaches infinity.

This is the limit of a quotient of functions. And we know that the limit of a quotient of functions is the quotient of their limits. So, we can find the limits of the numerator and denominator separately, should they exist. And as the limit of a sum of functions is the sum of their limits, we can find the limits term-by-term.

Now, we have lots of limits to evaluate but they’re all of very simple terms. And we can make them simpler by taking the constants outside the limits. As the limit of a constant times a function is that constant times the limit of the function. And now, the vast majority of our limits are of the form the limit of π‘₯ to the power of some negative number as π‘₯ approaches infinity.

What are the values of such limits? Well, we can write π‘₯ to the negative 𝑛 as one over π‘₯ to the 𝑛. And for 𝑛 greater than zero, the value is zero. All these limits then are zero. And we’re only left with the two limits to worry about, both of which are limits of the constants four and three. The limit of a constant function is just that constant. And so, being careful to include this minus sign, we see the answer is negative four over three.

Solving this problem was straightforward because we only had constants and negative powers in the numerator and denominator. And we know what the limit of a negative power of π‘₯ is as π‘₯ approaches infinity; it’s zero. Let’s now see an example where we don’t just have negative powers.

Find the limit of π‘₯ squared plus three all over eight π‘₯ cubed plus nine π‘₯ plus one as π‘₯ approaches infinity.

Our first thought might be to use the fact that the limit of a quotient is the quotient of the limits. This gives us the limit of π‘₯ squared plus three as π‘₯ approaches infinity over the limit of eight π‘₯ cubed plus nine π‘₯ plus one as π‘₯ approaches infinity. But we run into problems because neither limit is defined. In the numerator, as π‘₯ approaches infinity, π‘₯ squared plus three doesn’t approach any real value, it just gets bigger and bigger without bound. And the same thing happens in the denominator. As π‘₯ increases without bound, the cubic eight π‘₯ cubed plus nine π‘₯ plus one also increases without bound.

Or maybe you think both limits should be infinity. And so, the limit on left-hand side is infinity over infinity. But this, just like a zero over zero, is an indeterminate form. And it doesn’t tell us the value of our limit. We need to use a different approach.

The trick to this question is to find the highest power of π‘₯ that appears in the numerator or denominator. That’s π‘₯ cubed here. And having found this highest power, we divide both numerator and denominator by it. What do we get? π‘₯ squared divided by π‘₯ cubed is π‘₯ to the negative one. And three divided by π‘₯ cubed is three π‘₯ to the negative three. And in the denominator eight π‘₯ cubed divided by π‘₯ cubed is just eight. Nine π‘₯ divided by π‘₯ cubed is nine π‘₯ to the negative two. And one divided by π‘₯ cubed is π‘₯ to the negative three.

So, now, we just have negative powers of π‘₯ and a constant in the numerator and denominator. And as a result, when we apply this limit law, we find that the limits in the numerator and denominator do now exist. Let’s find their values. We can use the fact that the limit of a sum of functions is the sum of their limits. This allows us to find the limit of each term separately. We can also take the coefficient outside the limits.

And now, apart from one limit, which is the limit of a constant function, and whose value must therefore be eight, all the other limits have the form the limit of π‘₯ to the power of negative 𝑛 as π‘₯ approaches infinity. Where 𝑛 is, of course, greater than zero. And we know the value of such limits. The value is always zero.

So, this is zero, and this is zero, and this is zero, and this is zero. Simplifying then, our answer is zero over eight, which is, of course, just zero. Now, in the first failed attempt at solving this problem, we said that the limits of the numerator and denominator individually were both undefined, or infinite. What do we mean by saying that the value of such limits can be infinite. Let’s find out.

If the limit of 𝑓 of π‘₯, as π‘₯ approaches infinity, is infinity, this means that we can make the value of 𝑓 of π‘₯ arbitrarily large by choosing π‘₯ to be large enough. Suppose you want 𝑓 of π‘₯ to be greater than a billion. Well, there’s some value such that if we choose π‘₯ to be greater than that value, then 𝑓 of π‘₯ will be greater than a billion as required. Another way to think about this is that, beyond a certain point, as π‘₯ increases without bound, 𝑓 of π‘₯ also increases without bound.

Similarly, the limit of 𝑓 of π‘₯, as π‘₯ approaches infinity, being negative infinity means that as π‘₯ increases without bound, 𝑓 of π‘₯ decreases without bound. And for completeness, we write down the meanings when π‘₯ tends to negative infinity as well. Let’s see an example.

Find the limit of six π‘₯ squared over π‘₯ minus six as π‘₯ approaches infinity.

There are various ways to find this limit. One way is to look at the graph of 𝑦 equals six π‘₯ squared over π‘₯ minus six. It looks like, as π‘₯ increases without bound, six π‘₯ squared over π‘₯ minus six also increases without bound. As a result, we can say that this limit is infinity. But you might not be convinced by this. Perhaps, the graph does something slightly different further along the π‘₯-axis.

We can also perform a polynomial long division to find out six π‘₯ squared over π‘₯ minus six equals six π‘₯ plus 36 plus 216 over π‘₯ minus six. And it’s straightforward to take limits on the right-hand side. We can do this term-by-term. The limit of six π‘₯, as π‘₯ approaches infinity, must be infinity. As π‘₯ increases without bound, six π‘₯ also increases without bound. The limit of 36, as π‘₯ approaches infinity, is just 36. This is the limit of a constant.

And the last limit might be a bit more tricky. We divide numerator and denominator by the highest power of π‘₯ that we see; that’s π‘₯. The limit of a quotient is the quotient of the limits. And the limit in the numerator is just zero and in the denominator is just one. So, the value of this limit is zero. So, our limit is infinity plus 36. And when we’re dealing with limits, it’s perfectly fine to say that infinity plus 36 is just infinity, which gives another path to this answer. Okay, now, let’s see a final problem.

Find the limit of nine minus eight π‘₯ plus six π‘₯ squared minus two π‘₯ cubed as π‘₯ approaches negative infinity.

The first thing we might be tempted to do is to write this limit of a sum as the sum of some limits. We can then evaluate each of these limits one-by-one. The limit of the constant function nine is just nine. What can we say about the limit of eight π‘₯ as π‘₯ approaches negative infinity? Well, with the graph of 𝑦 equals eight π‘₯ in mind, we can see that as π‘₯ decreases without bound, 𝑦 also decreases without bound. And so, the limit of eight π‘₯, as π‘₯ approaches negative infinity, is negative infinity.

How about the limit of six π‘₯ squared as π‘₯ approaches negative infinity? Again, we have the graph in mind, and we see that as π‘₯ decreases without bound, 𝑦 increases without bound. So, this limit is infinity. And finally, the limit of two π‘₯ cubed as π‘₯ approaches negative infinity, we know what a cubic curve looks like. And we see that as π‘₯ approaches negative infinity, 𝑦 also approaches negative infinity. This limit is negative infinity.

So, it looks like our limit is nine minus negative infinity plus infinity minus negative infinity. And if we treat infinity like a number, we can write minus negative infinity as plus infinity, getting nine plus infinity plus infinity plus infinity. And this sum is equal to infinity. Now, we have to be slightly careful about manipulating infinity in this way. But it turns out that all of these steps are okay in this situation. We got lucky though that we didn’t have any minus signs left at the end, as infinity minus infinity is undefined.

For various reasons then, it might be worth seeing how to solve this problem in a different way. What we do instead is we factor out the highest power of π‘₯, that’s π‘₯ cubed, from inside the limit. This gives us the limit of π‘₯ cubed times nine π‘₯ to the negative three minus eight π‘₯ to the negative two plus six π‘₯ to the negative one minus two as π‘₯ approaches negative infinity. The limit of a product is the product of the limits.

Now, what’s the limit of π‘₯ cubed as π‘₯ approaches negative infinity? Well, we can slightly modify our graph and call this the graph of 𝑦 equals π‘₯ cubed instead. And we’ll see that this limit is negative infinity. How about this limit? Well, these terms with negative powers of π‘₯ contribute nothing, and so we’re left with just the limit of negative two as π‘₯ approaches negative infinity. And this is, of course, just negative two. The only bit of infinity manipulation we need to do is to multiply negative infinity by negative two. The minus signs cancel. And we get just infinity.

Alternatively, we could’ve factored the whole term negative two π‘₯ cubed out of the limit. And then, the value of the second limit in our product would just be one. We can easily show that the first limit in the product is infinity. And you might be more willing to believe that infinity times one is infinity than you were to believe that negative infinity times negative two is infinity.

Using this method, we can show that the limit of a polynomial, as π‘₯ approaches positive or negative infinity, is just the limit of the highest-degree term of that polynomial as π‘₯ approaches positive or negative infinity. Then, all we have to do is look at, or imagine, a graph of this monomial function.

Let’s see the key points we’ve covered in this video. We can consider limits of the form the limit of 𝑓 of π‘₯ as π‘₯ approaches positive or negative infinity. And in these cases, the limit laws still apply. The limit of the reciprocal function one over π‘₯, as π‘₯ approaches either positive or negative infinity, is zero. And hence, combining this with one of the limit laws, we see that the limit of one over π‘₯ to the power of 𝑛, as π‘₯ approaches positive or negative infinity, is also zero if 𝑛 is greater than zero.

We can find the limits of rational functions by dividing numerator and denominator by the highest power of π‘₯ and using the above result. And in a similar way, we can show that the limit of a polynomial, as π‘₯ approaches positive or negative infinity, is just the limit of its highest-degree term. We need to be slightly careful when playing with infinity. But with some exceptions, for example, the indeterminate forms infinity over infinity and infinity minus infinity, infinity can be manipulated like a real number in the context of limits.

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