For example:
$$\lim_{n\to\infty} n^{1/n} = \lim_{n\to\infty}e^{\ln n^{1/n}} = \lim_{n\to\infty}e^{1/n\ln n}= e^{\lim_{n\to\infty} 1/n \ln n}$$ $$ \overset{LH}{=} e^{\lim_{n\to\infty}1/n} = 1$$
But, more generally, if both functions are continuous:
$$\lim_{x\to\infty}(f\circ g)(x) \overset{?}{=}f\lim_{x\to\infty} g(x)$$
That works in the example, but really what I want to ask is whether this is a legitimate method if the limit is not known to converge. Is there a time when this method would produce an erroneous result?
The above comment essentially answers the question: the point is that continuity is not the key hypothesis (of course, it can be very strong) to establish a result about the limit of a composite function. The following is standard:
Let $f,g$ be two functions and $x_0$ be a limit point in the domain of $f \circ g $. If $\lim_{x \rightarrow x_0} g(x) = x_1 $, $\lim_{ x \rightarrow x_1} f(x) = a$ and there exists a neighbourhood $U$ of $x_1$ such that $g(x) \neq x_1 \ \ \forall x \in U \setminus \{ x_0 \}$ , then:
$$ \lim_{x \rightarrow x_0} (f \circ g ) (x) = \lim_{x \rightarrow x_0} f(g(x)) = a $$
We can add a continuity request to get a useful corollary of the above proposition:
In the same hypothesis, if $f$ is continuous at $x_1$ then:
$$ \lim_{x \rightarrow x_0} (f \circ g ) (x) = \lim_{x \rightarrow x_0} f(g(x)) = f(x_1) = f(\lim_{x \rightarrow x_0} g(x) ) $$