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:**

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

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

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*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