How do we show that the limit of $\frac{x^6-x^2\sin(\frac{1}{x^2})}{x^4}$ as $x$ tends to $0$ does not exist? I thought maybe we should consider two sequences that tend to 0 and show than $f(a_{n})$ and $f(b_{n})$ tend to different limits as n tends to infinity? But I cannot seem to find two sequences.
Thanks in advance
On
$$f(x):=\frac{x^6-x^2\sin\frac1{x^2}}{x^4}=x^2-\frac1{x^2}\sin\frac1{x^2}$$
Now, for example, choose the sequences
$$x_n=\frac1{\sqrt{2n\pi}}\;,\;\;y_n=\frac{\sqrt2}{\sqrt{(4n+1)\pi}}$$
Observe that both sequences tend to zero when $\;n\to \infty\;$ , so if the function's limit existed it'd be the same on both sequences, yet:
$$\begin{align}&f(x_n)=\frac1{2\pi n}-2\pi n\sin2\pi n=\frac1{2\pi n}\xrightarrow[n\to\infty]{}0\\{}\\&f(y_n)=\frac2{(4n+1)\pi}-\frac{(4n+1)\pi}2\sin\frac{(4n+1)\pi}2=\frac2{(4n+1)\pi}-\frac{(4n+1)\pi}2\xrightarrow[n\to\infty]{}\infty\end{align}$$
Hint For what values of $x$ does $\sin \left(\frac{1}{x^2}\right)$ achieves its maximum? Minimum?
(Alternatively, it suffices to prove that, say, the right-handed limit alone doesn't exist; it might be easier to see the limiting behavior by writing $x := \frac{1}{u}$ and considering the limit as $u \to +\infty$.)