Does $\ \sin \left(\frac{(k\pi)^2}{x+k\pi}\right)\ {\to \sin(x)}\ $ as odd integers $\ k { \to } \infty$? I am investigating $\sin(\frac{1}{x})$ and it's properties out of interest, and this question is related to my investigation.
Initially, I was trying to prove the result in the title for $0<x<2\pi$, but I do think I have shown that for all x, $\ \sin\left(\frac{(k\pi)^2}{x+k\pi}\right)\ {\to \sin(x)}\ $ as $\ k { \to } \infty,\ k\ $ is odd integers only, but just wanted to check my working is correct. I did the following:
\begin{align*}
  \sin\left(\frac{(k\pi)^2}{x+k\pi}\right)
    & = \sin\left(\frac{(x + k\pi)^2 - (x^2 + 2k\pi x)}{x+k\pi}\right) \\
    & = \sin\left(x+k\pi - \frac{x^2 + 2k\pi x}{x+k\pi}\right) \\
    & = \sin\left(x+k\pi - (x + \frac{k\pi x}{x+k\pi})\right) \\
    & = \sin \left(k\pi - \frac{k\pi x}{x+k\pi}\right)
  \end{align*}
Since $k\ $ is odd, this is equal to:
\begin{align*}
    & \sin(k\pi)\cos(\ldots) + \cos(k\pi)\sin \left(-\frac{-k\pi x}{x+k\pi}\right) \\
    = & \, 0 - \Bigl(-\sin \left(\frac{k\pi x}{x+k\pi}\right)\ \Bigr) \\
    = & \, \sin \left(\frac{k\pi x}{x+k\pi}\right) \\
    = & \, \sin \left(\frac{k\pi x + x^2 - x^2}{x+k\pi}\right) \\
    = & \, \sin \left( x - \frac{x^2}{x+k\pi}\right)
  \end{align*}
which, for all $\ x \in \mathbb R, \ { \to } \sin(x)\ $ as $ \ k { \to } \infty$
Firstly, is this all correct? And secondly have I done unnecessary steps or is there is a less convoluted way to get to the result?
Thanks in advance!

Note also that
\begin{align*}
  \sin\left(\frac{(k\pi)^2}{x+k\pi}\right)
    & = \sin\left(\frac{(k\pi)^2 - x^2}{x+k\pi} + \frac{x^2}{x+k\pi}\right) \\
    & = \sin\left(k\pi-x + \frac{x^2}{x+k\pi}\right) \\
    & = \sin\left(x - \frac{x^2}{x+k\pi}\right) \\
  \end{align*}
