Limits at infinity

Limits part 2: Functions: All kinds of limits

Limits at infinity

Just as one can study the long-term behaviour of sequences, one can also do that for functions. For example, the following holds: $\lim_{x\to\infty} \frac{1}{x}=0\text.$ The precise definition is:

The limit at infinity of a function $f : \mathbb R\to\mathbb R$ is $L$ if for every number $\epsilon > 0$ there exists a number $N$ such that if $x > N$ , then $|f(x) - L|<\epsilon$ .

This is denoted by $\lim_{x\to\infty} f(x) = L$ . Compare this with the definition of a limit of a sequence:

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

There is a subtle difference between these two definitions: in the definition of $\lim_{x\to\infty}f(x)=L$ must for all real numbers $x>N$ hold that $|f(x) - L|<\epsilon$ and in the definition of $\lim_{n\to\infty}a_n = L$ we consider only natural numbers $n > N$ . In the following example, this difference is emphasized:

Consider the function $f(x) = \sin(2\pi x)\text.$ The period of this function is $1$ , so: $f(0) = f(1) = f(2) = \dots = 0$ . Thus, when we look at the sequence $a_n = \sin(2\pi n)$ , we see that $a_n = 0$ all $n$ . So $\lim_{n\to\infty} a_n = 0$ , but the limit $\lim_{x\to\infty} f(x)$ does not exist because the function oscillates between $-1$ and $1$ .

Similarly, we define $\lim_{x\to-\infty} f(x)$ .

The limit at minus infinity of a function $f : \mathbb R\to\mathbb R$ is $L$ if for every number $\epsilon > 0$ there exists a number $N$ such that if $x < N$ then $|f(x) - L|<\epsilon$ .

$\lim_{x\to-\infty} \frac{1}{x}=0$

Let a number $\epsilon > 0$ be given and define $N = -\frac{1}{\epsilon}$ . If $x < N$ , then $0>\frac{1}{x} > \frac{1}{N} = -\epsilon$ because $N$ is negative. So we have $\frac{1}{x}\in (-\epsilon,0)$ and it follows that $\left| \frac{1}{x} - 0 \right| < \epsilon$ .