The Sequential Criterion for Limits at Infinity and Negative Infinity

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