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What would be $_{lim _{x→0}} [\frac{sinx}{x}]$ where $[.]$ denotes the greatest integer function.

According to me the answer should be 1 because of the result obtained from the sandwich theorem. That $_{lim_{x→0}} \frac{sinx}{x} = 1$. But everyone I've asked says that the answer is 0. But how can this be? I think it could be if the result $_{lim_{x→0}} \frac{sinx}{x} = 1$ is itself approximate.

Please let me know where I'm going wrong.

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    $\begingroup$ $\sin{x}/x < 1$ for $x \neq 0$. $\endgroup$
    – Chappers
    Apr 16, 2017 at 16:38
  • $\begingroup$ @fleablood One of the results that I knew was that lim $\frac{sinx}{x}$ as x tends to zero is 1. $\endgroup$ Apr 16, 2017 at 16:50
  • $\begingroup$ Okay, sorry. Brain fart. But you have $\sin x/x > [\sin x/x]$ Now you need a f(x) so that $f(x) < [\sin x /x]$ and $f(x) = 1$. YOu can't have a sandwhich with only a top piece of bread. . ANd Chappers and Stefon provve there is no bottom slice = 1. Ther's a bottom slice that equals 0 but that doesn't make a flat sandwich you can panini press. That makes a thick hero sandwich that never gets thin. 0 \le [\sin x/x] < 1$. That doesn't help. $\endgroup$
    – fleablood
    Apr 16, 2017 at 16:54

2 Answers 2

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Hint: For $x \in (-\pi, \pi) \setminus \{0\}$ you have

$$0<{\sin x \over x}<1. $$

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You can't have a sandwich with only one slice of bread. You know $[\frac {\sin x}{x} ] \le \frac {\sin x}{x} $ and $\lim \frac {\sin x}{x} = 1$. So $\lim [\frac {\sin x}{x} ] \le \lim \frac {\sin x}{x} =1$. But that isn't enough.

You need a lower slice of bread so $f(x) \le [\frac {\sin x}{x} ] $ and $\lim f(x) = 1$. There is no such $f$ as, apparently, $-1 \sin x/x < 1$ for $\-pi < x < \pi; x \ne 0$, it follows $ [\frac {\sin x}{x} ]= \{0, -1\}$. So $\lim [\frac {\sin x}{x} ] \le 0$ and you chose the wrong top slice of bread.

At this point we should just give up on the sandwich theorem.

If $0 < x \le \pi/2$, $0 < \sin x/x < 1$ and if $-\pi/2 \le x < 0$ then $-1 < \sin x/x < 0$ so $[\frac {\sin x}{x} ]_{\pi/2>x>0} = 0$ and $[\frac {\sin x}{x} ]_{-\pi/2< x < 0} = -l$ and $\lim_{x\rightarrow 0^+}[\frac {\sin x}{x} ] = 0$ and $\lim_{x\rightarrow 0^-}[\frac {\sin x}{x} ] =-1$.

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