The limit of 1/n

Limits part 1: Infinite sequences: Introduction

The limit of 1/n

The intuitive meaning of $\lim_{n\to\infty} a_n = L$ is that the numbers $a_n$ tend to $L$ as $n$ becomes very large. The more precise mathematical definition is as follows:

The limit of a sequence $(a_n)_{n=1}^\infty$ is $L$ if for all $\epsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_n - L|<\epsilon$ .

The $\epsilon$ in this definition one can view as a margin of error. If the margin of error is fixed, the definition states there exists an integer $N$ such that for all $n>N$ the element $a_n$ differs at most $\epsilon$ from $L$ . Hence, this definition states that for every margin of error you choose, the elements $a_n$ of the sequence eventually lie closer to $L$ than this margin of error.

In the exercise you saw that the limit of

$1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \tfrac{1}{5},\dots$ equals $0$ . In this animation you can see how quickly the sequence tends to 0.

$\displaystyle \lim_{n\to\infty} \frac{1}{n} = 0$ .

Suppose $\epsilon > 0$ is given. We are now looking for an $N$ such that for all $n>N$ the difference between $\tfrac 1n$ and 0 is less than $\epsilon$ . Choose $N = \lceil\tfrac{1}{\epsilon}\rceil$ , this means that we round up $\tfrac{1}{\epsilon}$ . For $n>N$ we now indeed have $\frac{1}{n} < \frac{1}{N}<\epsilon$ and it follows that $|a_n - 0| < \epsilon$ for all $n > N$ . This shows that $\lim_{n\to\infty} \tfrac 1n = 0$ .

From now on you can always use the limit $\displaystyle\lim_{n\to\infty} \frac{1}{n} = 0$ without proving it again. This limit can be used to calculate complicated limits. In the next section we will examine how this works.