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yes its 1

Hint :
for all \(x\in \mathbb{R}\)\[0\le x-\lfloor x \rfloor<1\]

it has to do something with squeeze theorem

thank u for ur effort :)

\[0\le \frac{1}{x}-\lfloor \frac{1}{x} \rfloor<1\]\[-\frac{1}{x}\le -\lfloor \frac{1}{x} \rfloor<1-\frac{1}{x}\]we're doin limit so x will not reach 0 so multiply both sides of later thing by x\[-1\le -x\lfloor \frac{1}{x} \rfloor

Nice! I think that works, and am glad to see how to do it:).