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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}

Answer

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 :)

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    $\begingroup$ How did you come up with $\lim_{x \to 0^-}\frac{\sin(1/x)}{1/x}=1$? $\endgroup$
    – Math Lover
    Sep 23, 2020 at 17:20
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    $\begingroup$ 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 $0.$ $\endgroup$
    – Dave L. Renfro
    Sep 23, 2020 at 17:24
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    $\begingroup$ 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 $1$. $\endgroup$
    – Ned
    Sep 23, 2020 at 17:25
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    $\begingroup$ 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. $\endgroup$
    – Dave L. Renfro
    Sep 23, 2020 at 17:29
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    $\begingroup$ Yes, $\theta$ can be anything as long as it goes to $0$ as well. $\endgroup$
    – Stefan Octavian
    Sep 23, 2020 at 17:31

1 Answer 1

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

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