This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

The Sequential Criterion for Limits at Infinity and Negative Infinity

Recall that $\lim_{x \to \infty} f(x) = L$ if $\forall \epsilon > 0$ $\exists M > 0$ such that if $x \in A$ and $x > M$ then $\mid f(x) - L \mid < \epsilon$. Similarly, $\lim_{x \to -\infty} f(x) = L$ if $\forall \epsilon > 0$ $\exists M < 0$ such that if $x \in A$ and $x < M$ then $\mid f(x) - L \mid < \epsilon$.

We will now look at what is known as The Sequential Criterion for Limits at Infinity and Negative Infinity, which is analogous to the other sequential criterion theorems.

Theorem 1: Let $f : A \to \mathbb{R}$ be a function and suppose that $(M, \infty) \subseteq A )$ for some $M \in \mathbb{R}$. Then $\lim_{x \to \infty} f(x) = L$ if and only if for every sequence $(a_n)$ from $A$ such that $\lim_{n \to \infty} a_n = \infty$ we have $\lim_{n \to \infty} f(a_n) = L$.

• Proof: $\Rightarrow$ Suppose that $\lim_{x \to \infty} f(x) = L$. Then $\forall \epsilon > 0$ $\exists M > 0$ such that if $x \in A$ and $x > M$ then $\mid f(x) - L \mid < \epsilon$. Let $(a_n)$ be a sequence in $A$, and let this sequence diverge to $\infty$, that is $\lim_{n \to \infty} a_n = \infty$. Now since $\lim_{n \to \infty} a_n = \infty$, then $\forall K > 0$ $\exists N \in \mathbb{N}$ such that if $n ≥ K$ then $a_n > M$, and so for $n ≥ K$ we must have that $a_n > M$ and so $\mid f(a_n) - L \mid < \epsilon$, and so $\lim_{n \to \infty} f(a_n) = L$.

• $\Leftarrow$ Suppose that for all sequences $(a_n)$ from $A$ such that $\lim_{n \to \infty} a_n = \infty$ we have that $\lim_{n \to \infty} f(a_n) = L$. We want to show that $\lim_{x \to \infty} f(x) = L$.

• Suppose not, that is, suppose that $\exists \epsilon_0 > 0$ such that $\forall M > 0$ then $\exists x_M > M$ such that $\mid f(x) - L \mid ≥ \epsilon_0$. Since $\lim_{n \to \infty} f(a_n) = L$ then $\forall \epsilon > 0$ $\exists N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid f(a_n) - L \mid < \epsilon$. But then this would not be true for $\epsilon_0$, a contradiction, so our assumption that $\lim_{x \to \infty} f(x) \neq L$ was false.

• Therefore $\lim_{x \to \infty} f(x) = L$. $\blacksquare$