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Find $$\lim_{x\to 0} [\frac{\sin{x}}{x}]$$ and $$\lim_{x\to 0} \{\frac{\sin{x}}{x}\},$$ if they exist, where $\{x\}$ is the fractional part and $[x]$ is the integer part of $x$.

Attempt

We have $[\frac{\sin{x}}{x}]+\{\frac{\sin{x}}{x}\}=\frac{\sin{x}}{x}$

we have to find $\lim_{x\to 0+} [\frac{\sin{x}}{x}]$ and $\lim_{x\to 0-} [\frac{\sin{x}}{x}]$

Also $\lim_{x\to 0+} \{\frac{\sin{x}}{x}\}$ and $\lim_{x\to 0-} \{\frac{\sin{x}}{x}\}$

Please help me to determine the above four limits.

  • 0
    Are you sure of that limit $\;n\to 0\;$? Perhaps it should be $\;\color{Red}x\to 0\;$ ? – 2017-02-27
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    @Crostul I need all four limits (Left and Right hand limits). Please help. – 2017-02-27
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    Edited $x\to 0$ – 2017-02-27

1 Answers 1

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Observe that for, say $\;x\in(-0.001,\,0.001)\setminus\{0\}\;$ , we have that $\;\frac{\sin x}x>0\;$ , and since the function is even it is enough to show for $\;x>0\;$, but then: we also know that $\;\sin x

$$\lim_{x\to 0^+}\frac{\sin x}x=1\implies \frac{\sin x}x<1\implies \left\lfloor\frac{\sin x}x\right\rfloor=0\xrightarrow[x\to0^+]{}0$$

and thus also

$$\left\{\frac{\sin x}x\right\}=\frac{\sin x}x\xrightarrow[x\to0]{}1$$

in the above punctured neighborhood.