Composition and Limits at Infinity It is a well known result that if a function $f$ is continuous at $b$ and $\lim_{x\rightarrow a} g(x)=b$, then $\lim_{x\rightarrow a} f(g(x))=f(b)$. 
When does this hold at infinity? If $f$ is continuous and $\lim_{x\rightarrow \infty} g(x)=b$, is it true that $\lim_{x\rightarrow \infty} f(g(x))=f(b)$?
 A: Yes. This is maybe easier to see if we phrase convergence in terms of sequences. $$\lim_{x\to\infty} g(x) = b$$
Is equivalent to the statement " For every sequence $\{a_n\}$ going to infinity, $\{g(a_n)\}$ converges to $b$", and similarly, $$\lim_{x\to b}f(x) = f(b)$$ is true only if for every sequence $\{b_n\}$ converging to $b$, $f(b_n)$ converges to $f(x)$.
Now, let $\{a_n\}$ be any sequence going to $\infty$. Then $g(a_n)$ is a sequence going to $b$, thus $f(g(a_n))$ is a sequence going to $f(b)$. So we have "For every sequence $\{a_n\}$ going to infinity $f(g(a_n))$ goes to $f(b)$". Which means
$$\lim_{x \to \infty} f(g(x)) = f(b)$$
Alternatively a straightforward $N-\delta-\epsilon$ argument: Fix $\epsilon > 0$. By continuity there exists $\delta$ such that that $|y_1 - b| < \delta$ implies $|f(y_1) - f(b)| < \epsilon$. By definition of limit, there is $N \in \mathbb{R}$ such that $x > N$ implies $|g(x) - b| < \delta$. Thus if we take this same $N$, then $x > N$ implies
$$|g(x) - b| < \delta$$
and thus
$$|f(g(x)) -f(b)| < \epsilon$$.
