I know that $x-1 < \lfloor x \rfloor \leq x$ hence $1-\dfrac{1}{x}<\dfrac{\lfloor x \rfloor}{x} \leq 1$ for $x>0$. I think it is easy to see from this that the limit should be 1 but I don't know how to formally prove this.
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5Do you know the squeeze theorem? – Eric Wofsey Nov 13 '16 at 22:54
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Comparison of limits. – egreg Nov 13 '16 at 22:55
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Heh, classic case of when not to use L'Hospital. – Simply Beautiful Art Nov 13 '16 at 22:59
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Is it me or is not the first time nor the second this exact question have been asked recently? – Nov 13 '16 at 23:00
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Ohh!! My apologizes, I didn't read the $x\to\infty$ part.... – Nov 13 '16 at 23:02
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4Duplicate: $\displaystyle\lim_{x\to \infty }\frac {\lfloor x \rfloor}{x}$. (Found using Approach0.xyz) – Workaholic Nov 14 '16 at 09:02
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You know that $1-\dfrac{1}{x}<\dfrac{\lfloor x \rfloor}{x} \leq 1$ for any $x > 0$ so taking the limits of the leftmost expression and of the rightmost expression you get:
$\lim_{x \to \infty} 1-\dfrac{1}{x} = 1$
and
$\lim_{x \to \infty} 1 = 1$
And since $\dfrac{\lfloor x \rfloor}{x}$ is always in between, by the squeeze theorem you get
$$\lim_{x \to \infty} 1-\dfrac{1}{x} \leq \lim_{x \to \infty} \dfrac{\lfloor x \rfloor}{x} \leq \lim_{x \to \infty} 1 \Rightarrow$$
$$\lim_{x \to \infty} \dfrac{\lfloor x \rfloor}{x} = 1$$
RGS
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