(Supposed) application of Squeeze Theorem

In my first calculus test of the semester, I had to evaluate the following limit:

$$\lim\limits_{x\to0^-}\frac {\sin^2(\frac 1 x)\sin^3x} {x} \ .$$

My working

$$\because \lim\limits_{\theta\to0^-}\frac {\sin\theta} {\theta} = 1$$

\begin{align} \therefore \lim\limits_{x\to0^-}\frac {\sin^2(\frac 1 x)\sin^3x} {x} & = \lim\limits_{x\to0^-}\frac {\sin^2(\frac 1 x)\sin^3x} {(\frac 1 x)(\frac 1 x)(x)(x)(x)} \\[5 mm] & = (1)(1)(1)(1)(1) \\[5 mm] & = 1 \end{align}

For $$x \in (-\frac \pi 2, 0) \ ,$$

$$0 \leq \sin^2(\frac 1 x) \leq 1$$ and $$x < \sin x < 0 \ .$$

$$\implies 0 \leq \frac {\sin^2(\frac 1 x)\sin^3x} x \leq x^2$$

Then, by Squeeze Theorem,

$$\lim\limits_{x\to0^-}\frac {\sin^2(\frac 1 x)\sin^3x} {x} = 0$$

I have two questions.

Firstly, I understand largely how the actual answer was derived, but the problem is I do not understand why my method did not work. If anyone can point out where I have gone wrong and why/how I went wrong, that will be great :)

Secondly, with regards to the actual answer, the only thing I still do not get is how the lower bound for $$x$$ was obtained. I understand that since we are taking the limit as $$x \rightarrow 0^-$$, the upper bound should be $$0$$, but I am not sure how the $$-\frac \pi 2$$ came about.

Edit

So following several comments from the community, it turns out this part of the question really was not that hard after all. On the bright side, I am probably never getting such a question wrong again :)

• How did you come up with $\lim_{x \to 0^-}\frac{\sin(1/x)}{1/x}=1$? Sep 23, 2020 at 17:20
• Note that the limit exhibited by @Math Lover is equivalent to $\lim_{y \rightarrow -\infty}\frac{\sin y}{y}$ (where $y = \frac{1}{x}),$ which is equal to $0.$ Sep 23, 2020 at 17:24
• You can reduce your method to the limit of $x^2 sin^2(1/x)$ which clearly goes to $0$ since the trig term oscillates between $0$ and $1$.
– Ned
Sep 23, 2020 at 17:25
• How did I make such a careless mistake. --- Actually, it took me at least 30 seconds to figure out what was wrong, because $\frac{\sin {(\text {stuff}})}{\text {stuff}}$ is so often used that one forgets to check that the appropriate limit operation is being applied. Sep 23, 2020 at 17:29
• Yes, $\theta$ can be anything as long as it goes to $0$ as well. Sep 23, 2020 at 17:31

$$\lim_{x\,\to\,0^-} \frac{\sin\frac 1 x}{\frac 1 x} = \lim_{u\,\to\,-\infty} \frac{\sin u} u = 0 \ne 1.$$