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