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.

Limits to Infinity and Negative Infinity

Definition: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then we say that the limit as $x \to c$ of $f$ is $\infty$ written $\lim_{x \to c} f(x) = \infty$ if $\forall \alpha \in \mathbb{R}$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $f(x) > \alpha$. Similarly, we say that the limit as $x \to c$ of $f$ is $-\infty$ written $\lim_{x \to c} f(x) = -\infty$ if $\forall \beta \in \mathbb{R}$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x -c \mid < \delta$ then $f(x) < \beta$.

The following diagrams represent the definition of a limit to infinity (left) and to negative infinity (right).

We will now look at a comparison theorem (analogous to that of sequences) of limits to infinity and negative infinity.

Theorem 1: Let $f : A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ be functions and let $c$ be a cluster point of $A$. Then: 1) If $\forall x \in A$, $x \neq c$ we have that $f(x) ≤ g(x)$ and $\lim_{x \to c} f(x) = \infty$ then $\lim_{x \to c} g(x) = \infty$. 2) If $\forall x \in A$, $x \neq c$ we have that $f(x) ≤ g(x)$ and $\lim_{x \to c} g(x) = -\infty$ then $\lim_{x \to c} f(x) = -\infty$.

• Proof: Suppose that $\lim_{x \to c} f(x) = \infty$ and let $\alpha \in \mathbb{R}$ be given. Then $\exists \delta > 0$ such that if $x \in A$, and $0 < \mid x - c \mid < \delta$ then $f(x) > \alpha$. But $g(x) ≥ f(x)$ $\forall x \in A$ and so for this $\delta > 0$, if $x \in A$ and $0 < \mid x - c \mid < \delta$ we have that $g(x) > \alpha$, and so $\lim_{x \to c} g(x) = \infty$.

• Similarly, suppose that $\lim_{x \to c} g(x) = -\infty$ and let $\beta \in \mathbb{R}$ be given. Then $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $g(x) < \beta$. But $f(x) ≤ g(x)$ $\forall x \in A$ and so for this $\delta > 0$, if $x \in A$ and $0 < \mid x - c \mid < \delta$ we have that $f(x) < \beta$, and so $\lim_{x \to c} f(x) = -\infty$. $\blacksquare$