Limits to Infinity

Limits as x approaches infinity can be tricky to think about. This is because infinity is not a number that x can ever be equal to. To evaluate a limit as x goes to infinity, we cannot just simply plug infinity in for x and see what we get. As a result, things like \(\mathbf{e^{\infty}}\) and \(\mathbf{\frac{1}{\infty}}\) don’t actually have a value.

So how can we deal with infinity?

Although infinity doesn’t have a specific value and can’t be plugged into functions, we can think about what will happen to a given function as x approaches infinity.

All this really means is that x is continually getting infinitely large. And as x gets bigger and bigger and bigger, what y value will our function get closer and closer to?

Let’s look at a few common examples and what they mean.

One Divided by Infinity

Like I said before, infinity is not a value. Therefore, \(\frac{1}{\infty}\) isn’t an actual number and doesn’t have a value. However, what we want to think about is what y value 1/x will approach as x goes to infinity. This is exactly what is being asked when we see: $$\lim_{x \to \infty} \frac{1}{x}$$

So let’s think about what happens to 1/x when we plug in bigger and bigger numbers for x.

x\(\mathbf{\frac{1}{x}}\)
11
100.1
1000.01
1,0000.001
10,0000.0001
100,0000.00001
1,000,0000.000001
10,000,0000.0000001
100,000,0000.00000001

So you can see in the table above that as x gets bigger and bigger, 1/x gets closer and closer to 0. Or in other words,

as x approaches infinity, 1/x approaches 0

We would write this mathematically as: $$\lim_{x \to \infty} \frac{1}{x} = 0$$

We can also see this graphically using Mathway. Notice in the graph below that as the x value goes toward infinity, you can see the y value getting closer to the y-axis (y=0).

limit as x goes to infinity of 1/x

Limits Going to Infinity

The other common example I mentioned is the limit as x goes to infinity of \(\mathbf{e^x}\). Or $$\lim_{x \to \infty} e^x$$

Again, it doesn’t really make sense to say that we can just plug infinity in for x and get \(\mathbf{e^{\infty}}\). This doesn’t actually have a value. This isn’t a number. Instead, we want to think about what y value \(\mathbf{e^x}\) goes toward as x goes to infinity. So let’s look at what happens as we raise e to a larger and larger power.

x\(\mathbf{e^x}\)
12.718
27.389
454.598
6403.429
82,980.958
1022,026.466
1002.688 * \(\mathbf{10^{43}}\)

So we can see here that \(\mathbf{e^x}\) starts giving us very large numbers quite quickly. And as we continue to plug in larger values for x, \(\mathbf{e^x}\) will continue to get bigger and bigger and bigger.

as x approaches infinity, \(\mathbf{e^x}\) approaches infinity

We would write this mathematically as: $$\lim_{x \to \infty} e^x = \infty$$

Rational Functions

Finding the limit as x approaches infinity of rational functions is a common limit you will run into. This is important because this is how you find horizontal asymptotes of rational functions. You are just looking to see what y value your function will get really close to (without touching that value) as your x goes to infinity.

What is a rational function?

A rational function is a function that is a fraction where the top and bottom of the fraction are polynomials. Basically this just means that the numerator and denominator of the fraction will be a sum of a handful of terms that are a constant times x raised up to some power. So it will look like this: $$f(x)=\frac{a_nx^n + a_{n-1}x^{n-1} + … + a_2x^2 + a_1x + a_0}{b_mx^m + b_{m-1}x^{m-1} + … + b_2x^2 + b_1x + b_0}$$

How do you take the limit of a rational function?

There are really only 3 cases you need to consider and the video above discusses these three cases as well. Any rational function will fall into one of these three categories, and each limit within each category will work out the same.

All you need to do is look at the degree of the polynomial on the top and bottom of the fraction. The degree of a polynomial is the highest power that x is being raised to. So for example, \(\mathbf{y=-4x^5+6x^2+x-12}\) is a polynomial of degree 5, because the highest power of x is 5.

Case 1: Degree of numerator is larger than degree of denominator

If the degree of the numerator is higher than the degree of the polynomial on the denominator, then the limit will go to infinity or negative infinity. This will only depend on the sign of the coefficient of the highest power x term on the numerator.

If Degree(P(x)) > Degree(Q(x)), then \(\mathbf{\lim\limits_{x \to \infty} \frac{P(x)}{Q(x)}= \pm \infty}\)

Example:

$$\lim_{x \to \infty} \frac{x^4-3x^2+x}{x^3-x+2}$$ $$=\lim_{x \to \infty} \frac{x^4-3x^2+x}{x^3-x+2} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}}$$ $$=\lim_{x \to \infty} \frac{\frac{x^4}{x^3}-\frac{3x^2}{x^3}+\frac{x}{x^3}}{\frac{x^3}{x^3}-\frac{x}{x^3}+\frac{2}{x^3}}$$ $$=\lim_{x \to \infty} \frac{x-\frac{3}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}+\frac{2}{x^3}}$$ $$= \frac{\lim\limits_{x \to \infty}x-\lim\limits_{x \to \infty}\frac{3}{x}+\lim\limits_{x \to \infty}\frac{1}{x^2}}{\lim\limits_{x \to \infty}1-\lim\limits_{x \to \infty}\frac{1}{x^2}+\lim\limits_{x \to \infty}\frac{2}{x^3}}$$ $$= \frac{\lim\limits_{x \to \infty}x-0+0}{1-0+0}$$ $$=\frac{\lim\limits_{x \to \infty}x}{1}$$ $$=\lim_{x \to \infty}x$$ $$=\infty$$

Case 2: Degree of numerator is smaller than degree of denominator

If the degree of the numerator is smaller than the degree of the denominator then the limit will go to 0.

If Degree(P(x)) < Degree(Q(x)), then \(\mathbf{\lim\limits_{x \to \infty} \frac{P(x)}{Q(x)}= 0}\)

Example:

$$\lim_{x \to \infty} \frac{16x^4}{0.0001x^5+18x}$$ $$=\lim_{x \to \infty} \frac{16x^4}{0.0001x^5+18x} \cdot \frac{\frac{1}{x^5}}{\frac{1}{x^5}}$$ $$=\lim_{x \to \infty} \frac{\frac{16x^4}{x^5}}{\frac{0.0001x^5}{x^5}+\frac{18x}{x^5}}$$ $$=\lim_{x \to \infty} \frac{\frac{16}{x}}{0.0001+\frac{18}{x^4}}$$ $$= \frac{\lim\limits_{x \to \infty}\frac{16}{x}}{\lim\limits_{x \to \infty}0.0001+\lim\limits_{x \to \infty}\frac{18}{x^4}}$$ $$= \frac{0}{0.0001+0}$$ $$=0$$

Case 3: Degree of numerator is equal to degree of denominator

If the degree of the numerator is equal to the degree of the denominator then the limit will be equal to the coefficient of the highest power x term in the numerator divided by the coefficient of the highest power x term in the denominator.

If Degree(P(x)) = Degree(Q(x)), then \(\mathbf{\lim\limits_{x \to \infty} \frac{P(x)}{Q(x)}= \frac{a}{b}} \ \) where a is the coefficient of the highest power x term in P(x), and b is the coefficient of the highest power x term in Q(x).

$$\lim_{x \to \infty} \frac{x^2+7}{-3x^2}$$ $$=\lim_{x \to \infty} \frac{x^2+7}{-3x^2} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}$$ $$=\lim_{x \to \infty} \frac{\frac{x^2}{x^2} + \frac{7}{x^2}}{\frac{-3x^2}{x^2}}$$ $$=\lim_{x \to \infty} \frac{1 + \frac{7}{x^2}}{-3}$$ $$= \frac{\lim\limits_{x \to \infty}1 + \lim\limits_{x \to \infty}\frac{7}{x^2}}{\lim\limits_{x \to \infty}-3}$$ $$= \frac{1+0}{-3}$$ $$=-\frac{1}{3}$$

Other Examples

\(\mathbf{\lim\limits_{x \to \infty} arctan \big( e^x \big)}\) | Solution

\(\mathbf{\lim\limits_{x \to \infty} \frac{sin(x)}{x}}\) | Solution

\(\mathbf{\lim\limits_{x \to \infty} \frac{ln(x)}{\sqrt{x}}}\) | Solution

Solutions

Example 1 Solution

\(\mathbf{\lim\limits_{x \to \infty} arctan \big( e^x \big)}\)

This is going to be based on the fact that we already discussed above that \(\mathbf{\lim\limits_{x \to \infty} e^x = \infty}\). Since we know that, we can say that \(\mathbf{y=e^x}\) and rewrite our limit as: $$\lim_{y \to \infty} arctan(y)$$

This is because y goes to infinity as x goes to infinity since \(\mathbf{y=e^x}\). Now we can find this limit by looking at a graph of y=arctan(x).

limit as x approaches infinity of arctan(x) or tan^{-1}(x)

Looking at this graph we can see that y=arctan(x) has a horizontal asymptote at \(\mathbf{y=\frac{\pi}{2}}\). As x goes toward infinity, you can see that the y value of our function gets closer and close to \(\mathbf{\frac{\pi}{2}}\). So this tells us $$\lim_{y \to \infty} arctan(y)=\frac{\pi}{2}$$ $$\lim_{x \to \infty} arctan \big( e^x \big)=\frac{\pi}{2}$$

Example 2 Solution

\(\mathbf{\lim\limits_{x \to \infty} \frac{sin(x)}{x}}\)

This one is going to use Squeeze Theorem. Click here to learn more about Squeeze Theorem and its required conditions if you don’t already know about them, then come back to this problem.

We know that \(\mathbf{-1 \leq sin(x) \leq 1}\) for all x. Since we are looking at this limit as x goes to positive infinity, we can also say that $$-\frac{1}{x} \leq \frac{sin(x)}{x} \leq \frac{1}{x}$$

We also know that \(\mathbf{\lim\limits_{x \to \infty} -\frac{1}{x}=0}\) and that \(\mathbf{\lim\limits_{x \to \infty} \frac{1}{x}=0}\). Since we know \(\mathbf{\frac{sin(x)}{x}}\) is between \(\mathbf{-\frac{1}{x}}\) and \(\mathbf{\frac{1}{x}}\) we can use Squeeze Theorem to say that $$\lim_{x \to \infty} \frac{sin(x)}{x}=0$$

Example 3 Solution

\(\mathbf{\lim\limits_{x \to \infty} \frac{ln(x)}{\sqrt{x}}}\)

For this limit, we will be able to use L’Hospital’s Rule. This is because ln(x) and \(\mathbf{\sqrt{x}}\) both go to infinity as x goes to infinity. This gives us an indeterminate form that is a possible application of L’Hospital’s Rule. We can also check to make sure that this meets the other conditions needed to apply L’Hospital’s Rule.

$$\lim_{x \to \infty} \frac{ln(x)}{\sqrt{x}}$$ $$=\lim_{x \to \infty} \frac{\frac{d}{dx}ln(x)}{\frac{d}{dx}\sqrt{x}}$$ $$=\lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2x^{1/2}}}$$ $$=\lim_{x \to \infty} \frac{1}{x} \cdot \frac{2x^{1/2}}{1}$$ $$=\lim_{x \to \infty} \frac{2}{x^{1/2}}$$ $$=\lim_{x \to \infty} \frac{2}{\sqrt{x}}$$ $$=0$$

Squeeze Theorem

The Squeeze Theorem is a useful tool for finding complex limits by comparing the limit to two much simpler limits. Squeeze Theorem tells us that if we know these three things:

$$1. \ \ \ g(x) \leq f(x) \leq h(x)$$

$$2. \ \ \ \lim_{x \to a} g(x) = L$$

$$3. \ \ \ \lim_{x \to a} h(x) = L$$

Then we also know that

$$\lim_{x \to a} f(x) = L$$

Keep in mind, requirement number 1 above only needs to be true around \(x=a\). It does not need to be true at \(x=a\) and it also does not need to be true when we get far away from \(x=a\). This is because limits only care what is happening to a function when we get infinitely close to a given \(x\) value, not at that \(x\) value and certainly not far away from that \(x\) value.

Example 1

You will typically see this used when you are finding the limit of a function which is made up of two functions being multiplied together and one of them is a trig function. For example, consider

$$\lim_{x \to 0}x ^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg).$$

At first glance, this limit looks terribly complicated. Especially because the function \(f(x) = x ^4 sin \Big( \frac{1}{\sqrt{x}} \Big)\) doesn’t even exist at \(x=0\) (plugging in \(x=0\) directly causes you to divide by \(0\), which can’t be done). However, if we use Squeeze Theorem, we can find this limit by finding two other limits that are much easier to find.

Squeeze Theorem says that we will need to find one function that is greater than or equal to \(f(x)\) around \(x=0\) (because we are finding the limit as \(x \to 0\)) and another function that is less than or equal to \(f(x)\) around \(x=0\). It is not always obvious which functions to use for comparison, but any function that is made up of \(sine\) or \(cosine\) of something multiplied by something else will usually follow the same pattern, so this is a good thing to try first.

What’s the pattern?

Consider the function \(y = sin(x)\). No matter what you plug in for \(x\) you will always get a \(y\) value between \(-1\) and \(1\). This will be true regardless of what you are taking the \(sine\) of. Even if you were to take \(sin\Big( \frac{1}{\sqrt{x}} \Big)\). No matter what you put in for \(x\) this will give you something between \(-1\) and \(1\), except if you try to put in \(x=0\) (remember you can’t divide by \(0\)).

But remember, with limits it doesn’t matter what happens at the \(x\) value we are approaching, only what is happening around that point. Therefore, it doesn’t matter that this function isn’t defined at \(x=0\) because it is defined for all \(x\) values near \(x=0\).

The point I’m trying to make is that \(sine\) of anything can be bound between \(-1\) and \(1\), so we can say:

$$-1 \leq sin(x) \leq 1, \ and$$

$$-1 \leq sin \bigg( \frac{1}{\sqrt{x}} \bigg) \leq 1$$

How does this relate to our example?

Now, the trick that makes this whole thing work is the fact that multiplying both sides of an inequality by a positive number maintains the inequality. Therefore, we can also multiply both sides of the inequality by a function that always outputs a positive number (or \(0\)).

For example, we can multiply both sides by \(x^4\), or all three “sides” in this case. We can do this because any number raised to an even power will always be positive (or \(0\)). This gives us

$$x^4 \cdot \Bigg[ -1 \leq sin \bigg( \frac{1}{\sqrt{x}} \bigg) \leq 1 \Bigg]$$

$$x^4 \cdot (-1) \ \leq \ x^4 \cdot \Bigg(sin \bigg( \frac{1}{\sqrt{x}} \bigg) \Bigg) \ \leq \ x^4 \cdot (1)$$

$$-x^4 \ \leq \ x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) \ \leq \ x^4$$

Now, by using the Squeeze Theorem, since \(x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg)\) is trapped between \(-x^4\) and \(x^4\) near \(0\), if we find that if

$$\lim_{x \to 0} -x^4 = \lim_{x \to 0} x^4$$

then that will also be the answer for the limit we are looking for. Consider the graph of \(y=-x^4\) and \(y=x^4\) around \(x=0\) shown below which was graphed using Desmos.

squeeze theorem

You can see that for both of these functions, as we travel along each function from both sides and get closer and closer to \(x=0\), we get closer and closer to a \(y\) value of \(0\). This tells us that

$$\lim_{x \to 0} -x ^4 = 0, \ and$$

$$\lim_{x \to 0} x ^4 = 0$$

Putting it all together

These two facts in combination with the inequality we showed earlier: $$-x^4 \ \leq \ x^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) \ \leq \ x^4,$$ tells us that

$$\lim_{x \to 0}x ^4 sin \bigg( \frac{1}{\sqrt{x}} \bigg) = 0.$$

Note: Compare these last 4 statements to the 4 statements at the beginning of this lesson.  The inequality is exactly the first thing I said we were looking to be able to show and the two limits are exactly like part 2 and 3.  So it’s no surprise that these three pieces combine to prove the limit we were trying to find.

Example 2

\(\mathbf{\lim\limits_{x \to \infty} \frac{sin(x)}{x}}\) | Solution

The Squeeze Theorem is an important application of limits in calculus, but I have written several other lessons and problem solutions about limits!  You should go check out my limits page to see what else I’ve done.  If I haven’t answered any questions you have on that page, let me know by sending me an email at jakesmathlessons@gmail.com and I’ll do my best to address your questions.

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