Does the Squeeze Theorem apply to $\lim_{x\to\infty}\sin(\frac{\pi x}{2-3x})$? I was recently asked to find the limit of 
$$\lim_{x\to\infty}\sin(\frac{\pi x}{2-3x}).$$
I eventually solved it by breaking the limit down to a composition of functions, but my first instinct was actually to invoke the Squeeze Theorem. My logic was that since 
$$-1\leq \sin(x) \leq 1,$$
I can have $\lim_{x\to\infty}-1$ serve as the lower bound and $\lim_{x\to\infty}1$ serve as the upper bound to conclude that
$$\lim_{x\to\infty}-1=\lim_{x\to\infty}1=\lim_{x\to\infty}\sin(\frac{\pi x}{2-3x})=1.$$
Given that the second method I did yielded an answer of $-\frac{\sqrt{3}}{2}$, I'm not sure where I went wrong in my implementation of the Squeeze Theorem. My suspicion is that a) I was wrong in thinking $\frac{\pi x}{2-3x}$ would be akin to simply another $x$ if both were entered into $\sin(x)$, or b) I needed to break down the limit into compositions before invoking the Squeeze Theorem.
Here are some better bounds to use for your squeeze theorem ($x&gt;2$):

$$-\frac{\sqrt3}2-\frac\pi{9x}&lt;\sin\left(\frac{\pi x}{2-3x}\right)&lt;-\frac{\sqrt3}2$$

As $x\to\infty$, we see that $\frac\pi{9x}\to0$, hence the limit is $-\frac{\sqrt3}2$.



These bounds were obtained by noting that:

$$0&lt;x&lt;\frac12\implies\frac{d^2}{dx^2}f(x)&gt;0$$

Where $f(x)=\sin\left(\frac\pi{2x-3}\right)$ is your function after $x\mapsto1/x$.

As $f(x)$ is concave up, it follows that for $x&gt;0$ and sufficiently close to $0$, we have

$$f(0)+f'(0)x&lt;f(x)&lt;f(0)$$

Whereupon I obtained the values $f(0)=-\frac{\sqrt3}2$ and $f'(0)=-\frac\pi9$.
I made a very simple mistake:
$$\lim_{x\to\infty}-1\stackrel{?}{=}\lim_{x\to\infty}1$$
$$-1\ne1$$
Therefore I can't use $-1$ and $1$ as my bounds for the Squeeze Theorem.

credits to thedilated for pointing this out to me in comments.