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:

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

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

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

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

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

**Here are some rules for the limits of sequences:**

$\begin{aligned} \lim_{n \to \infty}\left( a_n \pm b_n \right) &= \lim_{n \to \infty} a_n \pm \lim_{n \to \infty} b_n \\ \lim_{n \to \infty} (c\cdot a_n) &= c \cdot \lim_{n \to \infty} a_n \\ \lim_{n \to \infty} \left( a_n b_n \right) &= \left( \lim_{n \to \infty} a_n \right) \left(\lim_{n \to \infty} b_n \right) \\ \lim_{n \to \infty} \frac{a_n}{b_n} &= \frac{\lim_{n \to \infty} a_n}{\lim_{n \to \infty} b_n} \text{, as long as } \lim_{n \to \infty} b_n \neq 0 \end{aligned}$

Now, look at these examples:

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

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

$\frac{12n^2-3n}{n^2+5n+9} = \frac{12-\frac{3}{n}}{1+\frac{5}{n}+\frac{9}{n^2}}$

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

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

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

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

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

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

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:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply