NanoMath said:
Would it be okay to take sequence an = √2/√nπ . Then this sequence obviously converges to zero as n goes to infinity. But f(an) alternates between 1,0,-1,0,...
There is infinitely many points approaching zero:
$$\frac{1}{\sqrt{\pi}}, \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}}, \frac{1}{\sqrt{4\pi}}, \frac{1}{\sqrt{5 \pi}}, \frac{1}{\sqrt{6 \pi}}, ...$$
Group these into triplets:
$$\left( \frac{1}{\sqrt{\pi}}, \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}} \right), \left( \frac{1}{\sqrt{4\pi}}, \frac{1}{\sqrt{5 \pi}}, \frac{1}{\sqrt{6 \pi}} \right), ...$$
Notice for each triplet of points inside ##(0, \frac{1}{\sqrt{\pi}}]##, the function ##\text{sin} \left(\frac{1}{y^2} \right)## will oscillate from ##0## to ##-1##, then to ##0##, then to ##+1## and then back to ##0##.
The triplets need not be ordered as I've shown. You can group different triplets and the function will oscillate in the same fashion, but potentially in a different order. For example, using:
$$\left( \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}}, \frac{1}{\sqrt{4\pi}} \right)$$
Will produce a different sequence for the oscillation (0,1,0,-1,0), but the behavior will be the same for each triplet. Namely, ##\left| \text{sin} \left(\frac{1}{y^2} \right) \right| \leq 1##.