Convergence without equations Let $f:[1,+\infty)\to \mathbb R$ be a function such that 
$$\lim\limits_{x\to +\infty} f(x) = L$$
Show that the sequence $f(n)$ converges and $\lim \limits_{n \to\infty}f(n)=L$. Find an example illustrating that the converse is in general not true.
Im not really sure whats being asked how can I show convergence without knowing the equation or $\limsup$ or $\liminf$?
 A: This is not very difficult to show by definition.
By definition, we have that $\lim\limits_{x\to\infty} f(x) = L$ is equivalent with the statement:

For all $\epsilon > 0$ there exists a number $M>0$ such that $$x > M \implies |f(x)-L| < \epsilon$$

Here, $x$ is a real number. If we constrain $x=n$ to the set of natural numbers, then the implication
$$n > M \implies |f(n)-L| < \epsilon$$
still holds. However, in the definition of the limit of a sequence, $M$ is a natural number, but this is no problem because we can always find a natural number $M^*$ that is larger than $M$ and the inequality $n>M$ is satisfied for $n>M^*$. Therefore, there exists $M^*\in\mathbb N$ such that
$$n > M^* \implies |f(n)-L| < \epsilon$$
hence, by definition
$$\lim\limits_{n \to \infty} f(n) = L$$
To show that the converse does not hold, consider the function (that was suggested by @Alexandros)
$$f(x) = \begin{cases} 1, x \in \mathbb N \\ 0, \text{otherwise} \end{cases}$$
Clearly the function $f(n)$ for $n \in \mathbb N$ is constant, and thus it converges, but $f(x)$ for $x\in \mathbb R$ diverges, because the function keeps oscillating between $0$ and $1$ however large $x$ gets.

Your question is quite interesting, because there is an alternative definition (by Heine) of the limit of a function, where the limit of a function is defined in terms of the limit of a sequence. It is stated through the following theorem:

$\lim\limits_{x\to a} f(x)=A$ if and only if $\lim\limits_{n\to\infty}f(x_n)=A$ for every sequence $(x_n)_{n\geq 1}$, $x_n\neq a$ such that $x_n\to a$ when $n\to\infty$.

Notice that for this theorem to hold, the above mentioned conditions need to be satisfied for all such sequences $(x_n)_{n\geq 1}$. You can see that the function $f(x)$ I defined earlier does not satisfy these conditions, because the sequences $a_n=\left(n+\frac{1}{2}\right)_{n\geq1}$ and $b_n=(n)_{n\geq1}$ both $\to \infty$ as $n \to \infty$, but $$\lim\limits_{n\to\infty} f(a_n) = 0$$
$$\lim\limits_{n\to\infty} f(b_n) = 1$$
and $0\neq1$.
A: You are just being asked to consider the sequence $u_n = f(n)$ and show that 
$$
\lim_{x \to +\infty} f(x) = \lim_{n\to \infty} u_n.
$$
The figure below is an example with $f(x)=\frac{x^2}{x^2+1}$ and $u_n =\frac{n^2}{n^2+1}$, both converging to $1$.

