What is this problem asking? (limits of functions) I was asked to prove or find a counterexample of the statement "Suppose that $f: \mathbb{R}\to \mathbb{R}$, and let $x_k=f(\dfrac{1}{k})$. Then $\lim_{k\to\infty} f(x_k)=\lim_{t\to0} f(t)$." I'm not sure what the statement is saying. If anyone could explain it, it would be appreciated.

There are two different types of limits being used in the statement
$$\lim_{k\to\infty}f(x_k)=\lim_{t\to 0} f(t).$$
On the left we have a sequence: 
$$f(x_1), f(x_2), f(x_3), \ldots$$
to which we might apply the definition of $x_k$ to rewrite as
$$f(f(1)), f(f(1/2), f(f(1/3)), \ldots$$
The limit of this sequence is given by $L$ if for every real number $\epsilon > 0$, there exists a natural number $N$ such that for all $n > N$, we have $|f(f(1/n) − L| < \epsilon$.
On the right, we have the limit of a function, which is given by $K$ if for all $\epsilon>0$, there exists a $\delta>0$ such that for all $t\in\mathbb{R}$ satisfying $0<|t−0|<\delta$, the inequality $|f(t)−K|<\epsilon$ holds.
So the question is asking you to determine whether, for any arbitrary function $f\colon \mathbb{R}\to\mathbb{R}$, the limits $K$ and $L$ here must be equal.
For a concrete example of what's going on, consider the function $f(x)=x$. Then $f(f(1/k))=f(1/k)=1/k$, so the sequence is just
$$1,1/2,1/3,\ldots$$
and it is easy to show that the limit is zero  by taking $N\geq 1/\epsilon$.
On the other hand $\lim_{t\to 0} f(t)$ is also zero, since in the definition of the limit we can always just take $\delta=\epsilon$. So for this particular function, the statement holds. Does it hold for all others?
