Verify whether the set $S:= \{x\cos\left(\frac{1}{x^4}\right): x\in [-9,2]\}$ is bounded above.
I'm not sure what is the correct approach, I started, assuming there exists an upper bound $u$ and then I used the definition to get $x\cos\left(\frac{1}{x^4}\right) \leq u \hspace{10pt}\forall x\in [-9,2]$.
Then I realised that $0\in [-9,2]$ but $f(0)$ is undefined, where $f(x)= x\cos\left(\frac{1}{x^4}\right)$. This it seems that $S$ can't be bounded above, but this can't be rigorous proof.
In its present form, the question does not make sense - for $x\cos\left(\frac{1}{x^4}\right)$ is not defined at $x = 0$. One option is to define $x\cos\left(\frac{1}{x^4}\right) = 0$ at $x = 0$, and another is to simply forget about $x = 0$. I have discussed the second one below.
Let's talk about $S = \left\{x\cos\left(\frac{1}{x^4}\right) : x\in [-9,0) \cup (0,2]\right\}$ because $x\cos\left(\frac{1}{x^4}\right)$ is not defined at $x = 0$. Note that $$\left|x\cos\left(\frac{1}{x^4}\right)\right| \le |x|$$ when $x \ne 0$, and for $x\in [-9,0) \cup (0,2]$ we have $|x| \le 9$. Hence, $9$ is an upper bound for $S$. $S$ is bounded above.