Consider $$\lim_{x\rightarrow 0} x\sin(1/x).$$
A hasty application of the squeeze theorem might look like
\begin{align*} -1 \leq \sin(1/x) \leq 1\\ -x \leq x\sin(1/x) \leq x \end{align*}
and so since $\lim_{x\rightarrow 0} -x = \lim_{x\rightarrow 0} x = 0$ by the Squeeze Theorem,
$$\lim_{x\rightarrow 0} x\sin(1/x) = 0.$$
Now, this isn't a correct application of the Squeeze Theorem because
$$-x \leq x\sin(1/x) \leq x$$
only when $x \geq 0$.
Essentially what goes wrong here is by multiplying by $x$ across the inequality, the inequality flips for certain values in a neighborhood of $0$ and prevents one function from being the upper (lower) bound on the entire neighborhood. However, it doesn't seem to matter because the definition of the limit ensures that the left and right handed limits agree and $f(x)$ will be sandwiched all the same.
My question is the following: Is there a function that would fail using the above strategy? Namely, start with some inequality on some smaller piece of the function, apply some sequence of operations to all three sides of the inequality until you get the desired sandwiched function and then completely disregard the sign flips on the way to the conclusion?