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I need your help proving/disproving the existence of this limit:

$$ \lim_{x \to 0} (e^x-1) \left(\frac{1}{x}-\left\lfloor{\frac{1}{x}}\right\rfloor\right) $$

I really don't even know how to approach this, thought of splitting into $0^+$ and $0^-$ but I'm not sure how to proceed from there.

Thank you

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    you can perhaps use the sandwich or squeeze theorem. If you think about the 1/x - floor(1/x) it is a saw tooth function with values on [0,1] – 2017-01-06

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First recap that $$f(x) \to 0 \iff |f(x)| \to 0$$ so we consider the absolut value and get:

$$\left|(e^x - 1)\cdot\left(\frac{1}{x} - \left\lfloor\frac{1}{x}\right\rfloor\right)\right| = |(e^x - 1)|\cdot\left|\left(\frac{1}{x} - \left\lfloor\frac{1}{x}\right\rfloor\right)\right| \le |(e^x - 1)|$$ because the difference between a value and it's floored value is always less or equal to $1$.

Limit on both sides gives you the result.