Introductory Taylor Series Limit Proof Does there exist a function $f : \mathbb R \rightarrow \mathbb R$ for which: 
$$\lim_{\substack{x\in\mathbb R\\ x\to \infty}}f(x)\neq\lim_{\substack{n\in\mathbb{N}\\ n\to \infty}}f(n)?$$
I was thinking maybe a function that had a root at every natural number and oscillated towards infinity (kind of like $x\cdot \sin x$). It is okay if one limit DNE while the other does. All ideas are appreciated. Thanks!
 A: What about something like 
$$\exp(x^2\sin^2(\pi x)).$$
The limit as $n\to\infty$ exists, the function is $1$ at the integers. And the function blows up for some $x$. 
Remark: We used this instead of a simpler example because of the "Taylor series" in the title. 
A: You could let
$$
f(x)=
\begin{cases}
17 & x\not\in\mathbb Q \\
0 & x\in \mathbb Q
\end{cases}
$$
Then $\displaystyle\lim_{x\in\mathbb R,x\rightarrow\infty}f(x)$ DNE while $\displaystyle\lim_{x\in\mathbb N,x\rightarrow\infty}=0$.
A: No, there is not such a function.
If the limit exists for $x \rightarrow +\infty$, it needs to exist for all sequences $(x_n)_{n \in \mathbb{N}}$ which tend to $+\infty$, in particular the natural numbers. If you write down $\lim$ on both sides, you assume that both limits exist.
However, if the function does not have a limit at $\infty$ because there are multiple limit points, it's possible that one subset of sequences converges towards one limit point, and another subset converges to a different limit point.
