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I'm trying to find $$\lim _{x\to 0} ((e^x-1)\cdot \operatorname{frac}\left(\frac{1}{x}\right))$$ I thought maybe use squeeze theorem but then I have $$(e^x-1)\cdot \left(\:\left(\frac{1}{x}\right)-1\right)\le (e^x-1)\cdot \operatorname{frac}\left(\frac{1}{x}\right)\le (e^x-1)\cdot \:\frac{1}{x}$$

where $\operatorname{frac}(x)=x-\lfloor x\rfloor$.

p.s - I can't use L'Hopipal rule.

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    What is frac(x). is it integer parts. if so the limit is zero. – 2017-01-01
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    it is $$x - ⌊x⌋$$ – 2017-01-01

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The fractional part is always between $0$ and $1$, and $e^x-1\to0$ as $x\to 0$, so you have something between $0\cdot(e^x-1)$ and $1\cdot(e^x-1)$, and those both approach $0$.

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    I just realized that I wrote the wrong thing... – 2017-01-01
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    I just realized that I wrote the wrong thing... It shouls be $$\left(\frac{1}{x}\right)$$ – 2017-01-01
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    @kurt you really need to specify if that is equal to $frac(x)$ or $frac(\frac{1}{x})$ – 2017-01-01
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    @hugh its $$frac(1/x)$$. I modified the question. – 2017-01-01
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Hint

we have that

$$\forall X\in \Bbb R\;\; 0\leq X-\lfloor X\rfloor \leq 1$$

$$\implies \forall x\neq 0\;\; |f(x)|\leq |e^x-1|.$$

the limit when $x\to 0$ is zero.