Prove that if $\lim\limits_{x\to 0}f\big(x\big(\frac{1}{x}-\big\lfloor\frac{1}{x}\big\rfloor\big)\big)$, then $\lim\limits_{x\to 0}f(x)=0$ Prove that if $\lim\limits_{x\to 0}f\bigg(x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)\bigg)$, then $\lim\limits_{x\to 0}f(x)=0$  
My attempt:  
If I can show that $\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg) \to 1$ as $x\to 0$, then we are done.  
We know, $n\le \dfrac{1}{x}\le n+1$, so $\bigg\lfloor\dfrac{1}{x}\bigg\rfloor=n$.  
But, I cannot do anything more to it. Please help. Thank you.
 A: Use the Squeeze Theorem.  First note that 
$$\frac{1}{x}-1<\left\lfloor\frac{1}{x}\right\rfloor\le\frac{1}{x}$$
from which we obtain
$$ \left|x\right|\ge \left|x\right|\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\ge 0$$
Therefore, as $x\to 0$, the Squeeze Theorem guarantees that
$$\lim_{x\to 0} x\,\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)=0 $$
which implies 
$$\lim_{x\to 0}f\left(\,x\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=\lim_{x\to 0}f(x)$$
If $f$ is continuous, then 
$$\lim_{x \to 0} f\left(x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=f\left(\lim_{x \to 0} \left[x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right]\right)=f(0).
$$ 
A: $$\lim_{x\to 0} {x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)}=\lim_{x\to 0} 1-x\bigg\lfloor\dfrac{1}{x}\bigg\rfloor$$
But $$\lim_{x\to 0}x\bigg\lfloor\dfrac{1}{x}\bigg\rfloor =1$$(Use the sequeeze theorem and the fact that $$\frac{1}{x}-1<[\frac{1}{x}]\le\frac{1}{x}$$ Therefore $$\lim_{x\to 0} f\bigg({x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)\bigg)}=\lim_{t\to 0}f(t)$$
in the latter, if the original limit is zero, so is the limit in RHS.
