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Within this entry, $\overline{\mathbb{R}}$ will be used to refer to the extended real numbers.
Theorem Let $f \colon \mathbb{R} \to \mathbb{R}$ be a function. Then $\overline{f} \colon \overline{\mathbb{R}} \to \overline{\mathbb{R}}$ defined by
$\overline{f}(x)=\begin{cases} f(x) & { if } x \in \mathbb{R} \\ A & { if } x=\infty \\ B & { if } x=-\infty \end{cases}$
is continuous if and only if $f$ is continuous such that $\displaystyle \lim_{x \to \infty} f(x)=A$ and $\displaystyle \lim_{x \to -\infty} f(x)=B$ for some $A,B \in \overline{\mathbb{R}}$
Proof. Note that $\overline{f}$ is continuous if and only if $\displaystyle \lim_{x \to c} \overline{f}(x)=\overline{f}(c)$ for all $c \in \overline{\mathbb{R}}$ By defintion of $\overline{f}$ and the topology of $\overline{\mathbb{R}}$ $\displaystyle \lim_{x \to c} \overline{f}(x)=\displaystyle \lim_{x \to c} f(x)$ for all $c \in \overline{\mathbb{R}}$ Thus, $\overline{f}$ is continuous if and only if $\displaystyle \lim_{x \to c} f(x)=\overline{f}(c)$ for all $c \in \overline{\mathbb{R}}$ The latter condition is equivalent to the
hypotheses that $f$ is continuous on $\mathbb{R}$ $\displaystyle \lim_{x \to \infty}f(x)=A$ and $\displaystyle \lim_{x \to -\infty}f(x)=B$ 
Note that, without the universal assumption that $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$ necessity holds, but sufficiency does not. As a counterexample to sufficiency, consider the function $\overline{f} \colon \mathbb{R} \to \mathbb{R}$ defined by
$\overline{f}(x)=\begin{cases} \displaystyle \frac{1}{x^2} & { if } x \in \mathbb{R} \setminus \{0\} \\ \infty & { if } x=0 \\ 0 & { if } x=\pm \infty. \end{cases}$
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