Limit of a Sequence

Limit of a Sequence Video

“The limit does not exist!” Cady Heron’s Mathletes answer may have been correct in the classic teen comedy Mean Girls, but limits often do exist in math and today we’re going to look at the limits of sequences.

Let’s look at the simplest sequence we can think of to get started:



Now this is a very simple sequence! It’s the set of counting numbers. So the first term is 1, the second term is 2, the seventh term is 7, and so on. The \(n^{ th}\) term, therefore, must be \(n\). We could write a rule for this sequence as \(x_{n}=n\). As the term number rises, so does the value of that place in the sequence. To find the limit of a sequence, we need to think about what will happen as \(n\) gets bigger and bigger and approaches infinity. In this case, it’s obvious. The value of the sequence will also approach infinity. But since infinity isn’t really a number, this sequence does not really have a limit! In math language that means it diverges. Or we can simply say it is divergent. This simply means that the sequence does not converge on an actual number.

Other arithmetic sequences will also not converge on an actual number. Here’s a slightly more complicated one:



This is an arithmetic sequence because each consecutive term is the same number apart. In this case, each term goes down by 3. The rule for this sequence is \(a_{n}=-3n+10\). We can explore where this one is going by finding the next terms in the sequence \((-11, -14, -17, -20…)\) or we could find the thousandth term using our rule.



Or the millionth term:



It’s pretty clear where this one is going too. It’s heading towards negative infinity, which means that this sequence also diverges. In math notation we would write it like this:

\(\displaystyle\lim_{n\rightarrow \infty }(-3n+10)=-\infty\)


This looks fancy but it’s just denoting that we’re finding the limit of the expression as \(n\) approaches infinity. That little bit under the “lim” could actually be something else if we were dealing with functions instead of sequences, like \(\rightarrow 0\). With sequences we’re dealing with positive values for the term numbers so we use \(n\) and try to discover what happens as \(n\) gets infinitely big.

So what kind of sequence does converge on an actual number?

Let’s find \(\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n}\).

We can find the first several terms of this sequence by plugging in values for \(n\):

\(\left \{{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8}} \right \}\)


If we convert those fractions to decimals it looks like this:

\(\left \{ 1,0.5,0.\overline3, 0.25,0.2,0.1\overline6, 0.\overline1\overline3\overline2\overline8\overline5\overline7, 0.125\right \}\)


Our sequence is not heading towards infinity! Or even negative infinity! It’s getting smaller and smaller but staying positive. So what is it approaching? Let’s find the hundredth term. That would be \(\frac{1}{100}\), which is 0.01. Getting smaller still. The thousandth term is \(\frac{1}{1,000}\), which is 0.001. I think we can see where it’s headed. Zero! But here’s the thing. It will never actually get there. It will just get closer and closer to 0 without ever actually being 0. We can say it converges on 0. Now we have our first convergent sequence. And more importantly, we have an important building block. We now know that \(\frac{1}{n}\) has a limit of 0. And as we’re about to see, that’s very useful when finding the limit of other sequences.

Fortunately, there are some rules for sequences that we can use with what we already know to help us find the limit of more complicated sequences. There’s a rule for addition, subtraction, multiplication and division. In math notation they look like this:

Addition Rule\(\displaystyle\lim_{n\rightarrow\infty}(a_{n}+b_{n})=\lim_{n\rightarrow\infty}a_{n}+\lim_{n\rightarrow\infty}b_{n}\)
Subtraction Rule\(\displaystyle\lim_{n\rightarrow\infty}(a_{n}-b_{n})=\lim_{n\rightarrow\infty}a_{n}-\lim_{n\rightarrow\infty}b_{n}\)
Multiplication Rule\(\displaystyle\lim_{n\rightarrow\infty}(a_{n}b_{n})=\lim_{n\rightarrow\infty}a_{n}\lim_{n\rightarrow\infty}b_{n}\)
Division Rule\(\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{\displaystyle\lim_{n\rightarrow\infty}}{\displaystyle\lim_{n\rightarrow \infty}}if \lim_{n\rightarrow \infty}b_{n}\neq 0\)

These four rules simply mean that we can break apart our sequence rule, find the limits of the parts, and then put it back together.

Let’s apply the addition rule to a sequence to illustrate this idea.

Find \(\displaystyle\lim_{n\rightarrow\infty}\left ( \frac{1}{n}+2 \right )\).

Our sequence has a rule of \((\frac{1}{n}+2)\). To find the limit of the sequence, we can break it apart and find the \(\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n}\) and then find the \(\displaystyle\lim_{n\rightarrow\infty} 2\) and simply add the limits together.

\(\displaystyle\lim_{n\rightarrow\infty} (\frac{1}{n}+2)=\lim_{n\rightarrow\infty}\frac{1}{n}+\lim_{n\rightarrow\infty}2\)


We’ve already found the \(\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n}=0\). So we could say that the limit for the first term of our expression is 0. So \(\displaystyle\lim_{x\rightarrow \infty }a_{n}=0\). But what is the limit of 2? For any constant sequence, the limit is the constant. The sequence represented by \(b_{n}=2\) looks like this:



Since every term is 2, the limit is 2. This is kind of obvious since it has nowhere else to go. Anyway, since we know the limit of the two parts, we can add them together.

\(\displaystyle\lim_{x\rightarrow \infty } \left ( \frac{1}{n} +2\right )=0+2=2\)


Now let’s try one that lets us use the subtraction and multiplication rules.

Find \(\displaystyle\lim_{x\rightarrow \infty } \left (1-3n^{-1} \right )\).

Here we have subtraction so we can break up our sequence:

\(\displaystyle\lim_{n\rightarrow\infty}(1-3n^{-1})=\lim_{n\rightarrow \infty }1-\lim_{n\rightarrow \infty }3n^{-1}\)


The limit of 1 is 1, as we saw in the last problem. Now we need to find the limit of \(3n^{-1}\). We can break that part down using the multiplication rule.

\(\displaystyle\lim_{n\rightarrow\infty}(1-3n^{-1})=\lim_{n\rightarrow \infty } 1-(\lim_{n\rightarrow \infty }3\cdot \lim_{n\rightarrow \infty }n^{-1})\)


The \(\displaystyle\lim_{n\rightarrow \infty }3=3\). Once again, we have the limit of a constant. The \(\displaystyle\lim_{n\rightarrow \infty }n^{-1}\) we also know, though this problem hides it a little bit. \(n_{-1}\) is \(\frac{1}{n}\), which we know has a limit of 0. So after all that, here’s what we know:

\(\displaystyle\lim_{n\rightarrow \infty }(1-3n^{-1})=1-(3\cdot 0)=1\)


After finding our three parts we do some simple math to find our limit is 1. Let’s take a look at the first few terms of our sequence to see if this makes sense.

\(\left \{ -2, -\frac{1}{2}, 0,\frac{1}{4},\frac{2}{5},\frac{1}{2},\frac{4}{7},\frac{5}{8},\frac{2}{3} \right \}\)


It does look like it is climbing towards 1. Let’s find the 100th term in the sequence just to make sure it hasn’t gone over the top.



Looks like it’s still getting closer to 1. If we look at the 1,000th term, we’d get 0.997, so you can see that we’re getting closer and closer to 1 but will never actually reach it.

That’s it for our introduction to the limits of sequences. I hope this video was helpful. Thanks for watching and happy studying!


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by Mometrix Test Preparation | This Page Last Updated: January 31, 2023