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How can I study the limit of the following function? $$\begin{equation*} \lim_{x \rightarrow 0} \frac{[x]}{\sin x} \end{equation*}$$

Any hint will be appreciated!

Thanks!

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    $\begingroup$ what does $[x]$ mean? floor function? $\endgroup$ – Siong Thye Goh Nov 6 '17 at 7:39
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    $\begingroup$ @Siong Yes, it means the greatest integer function in many countries. That is the floor function. $\endgroup$ – samjoe Nov 6 '17 at 7:47
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    $\begingroup$ @samjoe thanks. I usually use $\lfloor \rfloor$ $\endgroup$ – Siong Thye Goh Nov 6 '17 at 8:47
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You need to evaluate both right and left hand limits here, as $[x]$ (floor function) is not continuous at $x = 0$:

  1. If $x \in [0,1)$, then $[x] = 0$. So the right hand limit is simply $0$.

  2. If $x \in (-1, 0)$, then $[x] = -1$. The left hand limit is:

$$\lim_{x\to 0^-} \frac{-1}{ \sin(x)} =\infty$$

From here we see the limit does not exist.

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For $0<x<1 $

$$ \frac{[x]}{\sin x} =0 $$ Hence

\begin{equation*} \lim_{x \rightarrow 0^+} \frac{[x]}{\sin x}=0 \end{equation*} Whereas for $-1<x<0$

$$ \frac{[x]}{\sin x} = \frac{-1}{\sin x} $$

Thus \begin{equation*} \lim_{x \rightarrow 0^-} \frac{[x]}{\sin x}=\infty \end{equation*} does not exist.

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