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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.

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    $\begingroup$ The limit of $-1$ ansd $1$ as $x$ approaches infinity are not equal! You cant use squeeze theorem here. $\endgroup$ – thedilated Jul 5 '17 at 0:12
  • $\begingroup$ OH I forgot about that! Should I delete the question or append your comment as an answer? $\endgroup$ – Brian Lee Jul 5 '17 at 0:16
  • $\begingroup$ Remember, OP, the idea of the squeeze theorem is that if you have $f(x) \le g(x) \le h(x)$ and $f(x), h(x) \to L$ as $x \to a$, then $g$ gets "squeezed" between the two of them and can't go above or below $L$ as $x \to a$, and therefore $g(x) \to L$ also. Since $-1$ and $1$ don't both approach a common $L$, you certainly can't use $-1 \le \sin(x) \le 1$ for this purpose. We use the squeeze theorem on $\frac{\sin(x)}{x}$ because $|\cos(x)| \le |\frac{\sin(x)}{x}| \le 1$ - which isn't like using it on $\sin(\frac{\pi x}{2-3x})$, where we don't have such inequalities. $\endgroup$ – Chris Jul 5 '17 at 0:19
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Here are some better bounds to use for your squeeze theorem ($x>2$):

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

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:

$$0<x<\frac12\implies\frac{d^2}{dx^2}f(x)>0$$

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 $x>0$ and sufficiently close to $0$, we have

$$f(0)+f'(0)x<f(x)<f(0)$$

Whereupon I obtained the values $f(0)=-\frac{\sqrt3}2$ and $f'(0)=-\frac\pi9$.

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Note that for $x \neq 0$ $$\frac{\pi x}{2-3x} = \frac{\pi}{\frac2x-3} \to \frac{\pi}{-3} = -\frac{\pi}{3}$$ as $x\to\infty$.

Since $\sin$ is continuous we therefore have $$\sin(\frac{\pi x}{2-3x}) \to \sin(-\frac{\pi}{3}) = -\frac{\sqrt3}{2}$$

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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.

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