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.

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

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