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.