Limits of Sequences


In a general sense, the limit of a sequence is a value that it approaches with arbitrary closeness. Stated precisely: \( \displaystyle \lim_{n \to \infty } \left\{ a_n \right\}=L \) means that for every \( \epsilon > 0 \) there exists a positive integer \( M \) so that:

If n>M, then anL<ϵ \text{If } n > M \text{, then } \left| a_n - L \right| < \epsilon


To take a simple example, let's find limn{1n}\displaystyle \lim_{n \to \infty }\left\{ \frac{1}{n} \right\} .

It's fairly obvious that as n n gets larger, the value of 1n \frac{1}{n} approaches 0, but how can we use the definition to help make that explicit?

Let M=1ϵ M = \lfloor \frac{1}{\epsilon} \rfloor . Then for any n>M n>M :

1n0<11/ϵ0ϵ. \left| \frac{1}{n} -0 \right| < \left| \frac{1}{\lfloor 1/\epsilon\rfloor} -0 \right| \leq \epsilon.

Here are some rules for the limits of sequences:

limn(an±bn)=limnan±limnbnlimn(can)=climnanlimn(anbn)=(limnan)(limnbn)limnanbn=limnanlimnbn, as long as limnbn0 \begin{aligned} \lim_{n \to \infty}\left( a_n \pm b_n \right) &= \lim_{n \to \infty} a_n \pm \lim_{n \to \infty} b_n \\ \lim_{n \to \infty} (c\cdot a_n) &= c \cdot \lim_{n \to \infty} a_n \\ \lim_{n \to \infty} \left( a_n b_n \right) &= \left( \lim_{n \to \infty} a_n \right) \left(\lim_{n \to \infty} b_n \right) \\ \lim_{n \to \infty} \frac{a_n}{b_n} &= \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} \text{, as long as } \lim_{n \to \infty} b_n \neq 0 \end{aligned}

Now, look at these examples:

What is limnan \displaystyle \lim_{n \to \infty} a_n if {an}=12n23nn2+5n+9\displaystyle \left\{ a_n\right\} = \frac{12n^2-3n}{n^2+5n+9} ?

First, we can re-express the limit by dividing both numerator and denominator by n2 n^2 :

12n23nn2+5n+9=123n1+5n+9n2 \frac{12n^2-3n}{n^2+5n+9} = \frac{12-\frac{3}{n}}{1+\frac{5}{n}+\frac{9}{n^2}}

Using the above rules, we can see that limnan=1201+0+0=12 \displaystyle \lim_{n \to \infty} a_n = \frac{12-0}{1+0+0} = 12 . _\square


Find limn31/n \displaystyle \lim_{n \to \infty} 3^{1/n} .

Since 1n+1<1n \frac{1}{n+1}< \frac{1}{n} , it is clear that 31/(n+1)<31/n 3^{1/(n+1)}<3^{1/n} , which means the sequence must be decreasing as n n increases. Further, it is apparent that 1n3 1^n \leq 3 for any value of n n , so it follows that 131/n 1 \leq 3^{1/n} . This gives us limn31/n=L \lim_{n \to \infty} 3^{1/n} = L such that:

1L31/n for any value of n>1. 1 \leq L \leq 3^{1/n} \text{ for any value of } n > 1.

The limit must then be 1 1 . To see why, assume that it is not. Then we have: 1nLn3 as n. 1^n \leq L^n \leq 3 \text{ as } n \to \infty.

Since this is clearly absurd for any L>1 L > 1 , we see that the limit must be 1 1 . _\square

Note by Arron Kau
6 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

can't we simply put y=1/x and take limit as y approaches zero

nishant singh - 6 years, 1 month ago

Log in to reply

Of course, but he's just showing a different way of looking/tackling it.

Milly Choochoo - 6 years ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...