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Evaluate $\lim_{x \to 0} (x\lfloor\frac{1}{x}\rfloor)$

I’m trying to solve it by using the squeeze theorem but I’m stuck.

I’m looking for a function $g(x)$ such that $g(x) \leq x\lfloor\frac{1}{x}\rfloor \leq x(\frac{1}{x}) = 1$

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Can anybody give me a hint?

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By definition,

$$\frac{1}{x} – 1 < \left\lfloor \frac{1}{x} \right\rfloor \leq \frac{1}{x}.$$

- If $x > 0$, this implies $1 – x \le x \lfloor \frac{1}{x} \rfloor \le 1$, so by squeezing you get $\lim_{x \to 0^+} x \lfloor 1/x \rfloor = 1$.
- If $x < 0$, the inequalities are reversed: $1 – x \ge x \lfloor \frac{1}{x} \rfloor \ge 1$ and again by squeezing it follows that $\lim_{x \to 0^-} x \lfloor 1/x \rfloor = 1$.

The RHS limit and the LHS limit exist and are equal, thus $\lim_{x \to 0} x \lfloor 1/x \rfloor = 1$.

Here is an approach. We recall the identity

$$ \lfloor y\rfloor = y – \{y\} $$

where $ \{y\} $ is the fractional part of $y$ which has the property $ \{y\}<1. $

Let $y=\frac{1}{x}$ so the limit becomes

$$ \lim_{y\to \infty} \frac{y – \{y\}}{y} = 1-\lim_{y\to \infty} \frac{\{y\}}{y}=1-0=1. $$

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