Difference between $\lim_{n \to + \infty} f(n)$ and $\lim_{x \to + \infty} f(x)$ Let $n \in \mathbb{N}$ and $x \in \mathbb{R}$. Let $f \in \textbf{C}^{0}(\mathbb{R},\mathbb{R})$ and suppose its limit exists in $+ \infty$.
I know that if $\lim_{x \to +\infty} f(x) = l$, therefore $\lim_{n \to +\infty} f(n) =l$ (it follows from the the sequential characterization of limits). But is it true that if $\lim_{n \to +\infty} f(n) =l$ therefore $\lim_{x \to +\infty} f(x) = l$? What if $f$ isn't supposed continuous anymore?
The reverse implication is not true even if $f$ is continuous. Consider
$$f(x) =\cos 2\pi x$$
Then the integer limit is $1$ but there's no limit when you quantify over all reals.
The statement $\lim_{n \to +\infty} f(n) =l\implies\lim_{x \to +\infty} f(x) = l$ is not true in general, when it holds we need some bounding trick to prove it. 

As a classical example from

$$a_n=\left(1+\frac1n\right)^n \to e$$

since for any real $x\to \infty, \quad \exists n \quad x\in[n,n+1)$, then we conclude that also  $f(x)=\left(1+\frac1x\right)^x\to e$ indeed

$$\left(1+\frac1{n+1}\right)^n \le \left(1+\frac1x\right)^x\le \left(1+\frac1n\right)^{n+1}$$

and both RHS and LHS tends to $e$.