A strange way to prove that $\lim_{x \to \infty} \sin(\frac{x^3}{x^2+1})$ does not exist. Instead of trying to replace the $x$ in the fraction with two sequence that will take $\sin(x)$ to two different values, one can notice that $\frac{x^3}{x^2+1}$ is onto $\mathbb{R}$ (surjective). Hence we can chose any sequence in $\mathbb{R}$ and it will be a subsequence (or subfunction, I don't really know how to call it) of $\frac{x^3}{x^2+1}$. Meaning that if I take $u_k = \frac{\pi}{2} + 2k\pi$ (these values will make $\sin$ equal to 1) such that $k \in \mathbb{N}$ and $v_k = \frac{\pi}{2} + (2k+1)\pi$ (these values will make $\sin$ equal to -1), the equations $\frac{x^3}{x^2+1} = u_{k_0}$ and $\frac{x^3}{x^2+1} = v_{k_0}$ will always have a solution by the surjectivity of $\frac{x^3}{x^2+1}$.
So I can take the two sequence $u_k$ and $v_k$ and use them as follows:
$\lim_{k \to \infty} \sin(u_k) = 1$ and $\lim_{k \to \infty} \sin(v_k) = -1$ ($\sin$ here is constant in both cases, but the sequences $u_k$ and $v_k$ go to infinity). Hence  $\sin(\frac{x^3}{x^2+1})$ diverges.
Is this reasoning correct?
 A: $
\newcommand{\N}{\mathbb{N}}
\newcommand{\seq}[1]{\left( #1_n \right)_{n \in \N}}
\newcommand{\R}{\mathbb{R}}
$Essentially, though it's a bit cluttered. You can clean it up as follows:

Choose the sequences $\seq x,\seq y$ in $\R$ defined by the rule
$$x_n := \frac \pi 2 + 2n\pi \qquad y_n := \frac \pi 2 + (2n+1) \pi $$
Note that $f:\R \to \R$ given by $f(x) := x^3/(x^2+1)$ is an odd function, hence surjective. Consequently we may choose sequences $\seq{x'},\seq{y'}$ where
$$f(x_n') = x_n \qquad f(y_n') = y_n$$
Notice furthermore that $f(x_n'),f(y_n') \to \infty$ as $n \to \infty$.
We then observe that
$$\lim_{n \to \infty} \sin(f(x_n')) = \lim_{n \to \infty} \sin \left( \frac \pi 2 + 2n\pi \right) = 1$$
whereas
$$\lim_{n \to \infty} \sin(f(y_n')) = \lim_{n \to \infty} \sin \left( \frac \pi 2 + (2n+1)\pi \right) = -1$$
Hence, the limit of $\sin(f(x))$ as $x \to \infty$ does not exist, because there are sequences of points we can choose in the image of $\sin(f(x))$ which converge to different values.
