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?

  • $\begingroup$ I am not sure I understand exactly what kind of functions you are looking for. Wouldn't $f(x)=\lfloor x\rfloor \sin(1/x)$ be an example? $\endgroup$
    – Taladris
    Commented Sep 15, 2021 at 3:55
  • $\begingroup$ Not an answer to the question. But for the example, you should just be using $-|x|$ and $|x|$ as the bounds. $\endgroup$
    – ndhanson3
    Commented Sep 15, 2021 at 3:56
  • $\begingroup$ @Taladris That is close, but not quite what I was hoping to find because following the naive strategy, in the limit one would get $-1 \leq \lfloor x \rfloor \sin(1/x) \leq 0$ which would be inconclusive rather than generating an incorrect conclusion. $\endgroup$
    – JessicaK
    Commented Sep 15, 2021 at 4:10

1 Answer 1


If you're specifically trying to prove that $\lim_{x \to 0} f(x)g(x)=0$ by showing $-h(x)\le f(x) \le h(x)$, so $-g(x)h(x)\le f(x)\le g(x)h(x)$, and by then showing $\lim_{x \to 0}g(x)h(x)=0$, then no, ignoring the minus signs in your proof will never let you prove something untrue. That's because your proof could be rewritten to say $|f(x)g(x)|\le |g(x)h(x)|$. But the fact that you're actually proving $|f(x)g(x)|\le |g(x)h(x)|$ is the only reason that it's safe to ignore the minus signs. You aren't really ignoring the minus signs if you have to check that it's safe to ignore the minus signs.


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