Learn the Intuition and Simple Techniques to Solve Them

What are limits at infinity? These are limits where the independent variable x approaches infinity.

This is an exciting moment, probably for the first time you'll be dealing with infinity...

Now, what it means that x approaches infinity? In practical terms, it means that below the word limit you have x→ ∞ instead of x → a.

You probably are already familiar with the symbol for infinity, ∞.

Infinity is not a number, is more like an auxiliary concept that we use in the context of limits. When we say that x approaches infinity we mean that the variable x grows without bounds.

Or another way to put it is that x takes values greater than any number you can come up with. You say 10 million? The variable x is taking values greater than that.

Our concept of limit allows us to talk about these things precisely. For example, let's consider the following limit:

This is read "the limit as x approaches infinity of one over x".

Here you can't simply "plug" infinity and see what you get, because ∞ is not a number. However, we can guess what this limit will be using our intuitive understanding.

Take your calculator and try to divide 1 by a very big number. Now try to divide 1 by an even bigger number. You get very small numbers, right?

This means that 1 divided by x approaches 0 when x approaches infinity. Let me show you the graph of this function:

Now, to be a little strict, we need to specify whether x is approaching positive or negative infinity:

- x→ +∞ means that x is approaching big positive numbers. For example: 10 million, 50 million, etc.
- x→ -∞ means that x is approaching "big" negative numbers. For example, -10 million, -50 million, etc.
- x→ ∞ (without sign) means that x is taking big numbers, either positive or negative

There is a confusing convention of simply using x→ ∞ in any case. To be precise, to say that the limit when x→ ∞ is equal to something, means that the limits when x→ +∞ and x→ -∞ are equal to that something.

This is the case in the example of the function 1 over x.

In the graph above we can see that when x approaches very big numbers, either positive or negative, 1 divided by x approaches zero. So:

This means that the two limits, when x→ +∞ and when x→ -∞, are equal to zero. You'll get used to this notation with some more examples.

Solving Limits at Infinity

The neat thing about limits at infinity is that using a single technique you'll be able to solve almost any limit of this type.

In the following video I go through the technique and I show one example using the technique. In the text I go through the same example, so you can choose to watch the video or read the page, I recommend you to do both.

Let's look at this example:

We cannot plug infinity and we cannot factor. So, now we'll use **the basic technique used to solve almost any limit at infinity**. It is a little algebraic trick.

Remember the property of fractions that said that you can divide both the numerator and denominator by the same number and the fraction remains the same? It's just about that.

We will divide this limit by x to the greatest exponent found in the function. In this case it is xNow, we can divide each term by x³:

We can cancel out some things:

Now, we know that any number divided by a very very big number is equal (almost) zero. So, as x approaches infinity, all the numbers divided by x to any power will approach zero. Plugging this we have:

Other Techniques for Solving Limits at Infinity

In the following examples we won't be using the basic technique of dividing by the greatest power of x. We'll be using something even more basic. In the video I go through the same examples as in the text, so you can choose to watch and listen or read.

Suppose we have this limit:

We can "separate" this fraction:

We did the reverse of adding fractions. Try to add the two fractions in the right side and you'll get the original function. Whenever you have two or more terms in the numerator, and only one term in the denominator, you may try to do this.

You simply put each term in the numerator divided by the denominator and add them.

Now, we know that 1/x approaches zero when x approaches infinity. So, we have:

That was straightforward!

Now, we will look at a really interesting problem. Let's consider the limit:

In the numerator we have the sum of all numbers from 1 to "n", where "n" can be any natural number.

Now, as n approaches infinity, the number of terms in the numerator also approach infinity, because there are n terms. So, the numerator approaches an infinite sum. Isn't this interesting?

To solve this limit, let's try to remember some basic facts about arithmetic progressions.

An arithmetic progression is an ordered set of numbers, where there is a constant "difference" between successive terms.

For example: 1, 2, 3, 4,... This is an arithmetic progression. The difference between successive terms is 1. We also know the formula that gives us the sum of "n" terms of an arithmetic progression:

In the video above I show a short deduction of this formula. In our limit we have an arithmetic progression in the numerator. We have:

So, according to the formula we have:

We can replace this in our limit:

This equals:

Now, we can use the technique we used in the previous example. We have only one term in the denominator, so we will "separate" the fraction. We have:

And finally we have:

Wow! These type of results usually blow my mind. We found that the sum of all natural numbers up to n, divided by n², approach a finite value: one half.

Some limits at infinity may not exist. For example, let's try to calculate this limit:

We will use the basic technique of dividing by the greatest power of x. Let's divide all terms by x squared:

We have:

All numbers divided by any power of x will approach 0 as x approaches infinity. So, we have:

Division by zero is undefined, so this limit does not exist. Some authors of textbooks say that this limit equals infinity, and that means this function grows without bound.

This is intuitive, because as you divide 1 by very very small numbers, you get very big numbers.

Now let's turn our attention to limits at infinity of functions involving radicals. In this case we can also use the basic technique of dividing by x to the greatest exponent.

In the video I show the same example, so you can watch the video or read the rest of the page.

Let's consider the limit:

We have a radical in the numerator. We use the basic technique of dividing both the numerator and denominator. In this case we divide by x:

Remember that x equals the square root of x squared. So, we will insert the x in the numerator inside the radical. To do this we need to square it. So, we have:

Note that you can simply take x squared out of the square root and you'll have the original expression. Now, we divide each term:

We can simplify this to:

Now, again, all the terms divided by x will approach zero. So we have:

Here we have a situation we didn't have before. We can have either a positive or negative sign. This depends on whether x approaches positive or negative infinity. We can see this in the graph:

When x approaches positive infinity, the function approaches positive 1. And when x approaches negative infinity, the function approaches negative 1. The deduction of these two cases is explained with more detail in he video above.

**Day 7 of the Intuitive Online Calculus Course:***Limits at Infinity*- Solving Limits:
*Intuition and Examples* - The Squeeze Theorem:
*Intuition and Examples* - The Formal Limit Definition:
*Getting the Intuition Behind It*

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**Open Question: Find the Asymptotes of this Function**

Find the horizontal and vertical and oblique asymptotes of f(x):
To answer this question, leave a comment below.

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**Limit With Radicals**

Hi,
I've been trying to solve this limit for some time now. I tried the techniques you showed here but none seemed to work. I don't have a clue of how …

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**Limits to infinity of fractions with trig functions** Not rated yet

The problem is as follows:
d(t)= 100 / **8+4sin(t)**
Find the limit as t goes to infinity.
It is assumed that t>0.
How would this work?
I know …

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