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I've tried plotting the graph of the function in Desmos and WolframAlpha, and true to the limit, the function tends to 1.

I would assume by the product rule of limits, that $\lim_{x \to 0} x = 0$, and hence the whole lot should be zero, too, but apparently not, as the behaviour of the floor function somehow makes the function tend to 1.

Or does it behave like $\lim_{x \to 0} \frac{x}{x} = 1 \mid x \in \mathbb{R}\backslash \{0\}$—which is much more straightforward to deduce? If so, why? I'm a little confused.

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4 Answers 4

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A useful property of the floor function is that $$ z - 1 < \lfloor z \rfloor \le z $$ for all $z$. If we apply that here with $z = \frac{1}{x}$, we get $$ \frac{1}{x} - 1 < \left\lfloor\frac{1}{x}\right\rfloor \le \frac{1}{x}, $$ then multiplying by $x$ (assuming $x > 0$) $$ 1 - x < x \cdot \left\lfloor\frac{1}{x}\right\rfloor \le 1. $$ Now you see that it is bounded on either side by $1-x$ and $1$, and both of them approach $1$, so we can apply the squeeze theorem to get that the limit is $1$.

Try the $x < 0$ case, it's similar!

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    $\begingroup$ Thank you very much—that was clear and succinct. The question gave the hint you mentioned, but I completely forgot about the Squeeze Theorem. $\endgroup$
    – SRSR333
    Sep 26, 2018 at 14:39
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Other answers have shown how you can calculate the limit, I just want to focus on one comment you made:

I would assume by the product rule of limits, that $\lim_{x \to 0} x = 0$

The rule of limits does not apply here, because the rule of limits, i.e. that $$\lim_{x\to 0} f(x)g(x) = \lim_{x\to0} f(x)\cdot \lim_{x\to0}g(x)$$ only applies when both limits on the right exist.

In your case, the limit

$$\lim_{x\to 0}\lfloor\frac1x\rfloor$$ does not exist, so the rule of limits cannot be applied.

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If you write $x=\frac1y$ then $x\lfloor\frac1x\rfloor = \frac{\lfloor y\rfloor}{y}$. As $y-1\le \lfloor y\rfloor\le y$, the squeeze theorem shows $\lim_{y\to\infty} \frac{\lfloor y\rfloor}{y}=1$. This shows your limit is also $1$.

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I think I see the source of your confusion. The product rule for limits states that $$\lim_{x\rightarrow a}(f(x)g(x)) = \lim_{x\rightarrow a}f(x)\cdot \lim_{x\rightarrow a}g(x)$$ if both the limits on the right-hand side exist. But here, with $g(x)=\lfloor\frac{1}{x}\rfloor$, the second limit does not exist. So you can't use the product rule.

Note that if one of those limits exists and the other doesn't, then the limit of the products doesn't exist, unless the limit that exists is zero. In this case, the limit of the products can exist, as in your example; but we can't say whether that limit exists, or what its value is, without further investigation.

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