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I'm trying to find the point-wise limit of this function on the interval $(0, \infty)$ (I know it exists because it is a homework question). I'm having some difficulty with the ceilings. Clearly, whenever x is a natural number, $f_n = 1$, but I am stuck on what to do in other cases. I've tried decoding the ceiling as an infimum (i.e. $\lceil t \rceil = \inf \{m \in {N} : m \geq t\}$, but it hasn't really gotten me farther. Any ideas on how to proceed?

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Hint. For any fixed $x$ we have $$\frac{nx}{n\lceil x\rceil}\le f_n(x)< \frac{nx+1}{n\lceil x\rceil}\ .$$Now let $n\to \infty$.

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  • $\begingroup$ Awesome...I didn't even think about using the squeeze theorem. Thanks! $\endgroup$ – Andrew Martin Mar 10 '14 at 6:03
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You always have $nx \le \lceil nx \rceil < nx+1$, so dividing by $n$ gives $x \le { \lceil nx \rceil \over n} < x+{1 \over n}$. It follows that $\lim_n { \lceil nx \rceil \over n} = x$.

Hence $f_n(x) \to { x \over \lceil x \rceil}$.

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