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I have a slight problem computing $\lim_{n\to\infty} [\frac{n-1}{n}]$. ( $[.]$ denotes the floor fonction). I know that the limit cannot "go inside" as the fonction is not continuous in the integers. I have attempted to use the inequality: $[x] \leq x < [x]+1$ but that was also unsuccessful.

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    Is this $\frac{n-1}{n}$ or $n-\frac{1}{n}$ in the floor? In the former case, the value of the floor function is $0$ while in the later case, the value of the floor function is $n-1$. – 2017-01-04
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    [Question edited] – 2017-01-04

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For $n>1$, $[n-\frac{1}{n}]=n-1$. Thus the limit is $\infty$.

Now that you have changed the question, one should note that $$ \biggr[\frac{n-1}{n}\biggr]=0 $$ for all $n>1$, and the limit is $0$.

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    Thanks for your answer. However, it seems slightly unintuitive to me: we have [(n-1)/n] never reaching 1 when taking limits to infinity (whereas (n-1)/n tends to 1). I know that the two are not the same as the limits cannot commute, but is this answer just supposed to be accepted, or is there a more rigorous explanation? – 2017-01-04
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    I don't understand your comment. Are you asking why the limit and the floor function do not commute? – 2017-01-04
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    Let $f(x)=[x]$ and denote $a_n=\frac{n-1}{n}$. Then $\lim_{n\to\infty}a_n=1$ but $\lim_{n\to\infty}f(a_n)=0$ since $f(a_n)=0$ for every $n>1$. What is the confusion?? – 2017-01-04
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    Ok, I get it now. Thanks again. – 2017-01-04
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$$\left [ n - \frac{1}{n} \right ] > n - \frac{1}{n} - 1 \longrightarrow \infty$$

Now since you've changed the question : $$\left [ \frac{n-1}{n}\right ] = \left [1-\frac{1}{n}\right ] = 0$$

Thus the limit is 0.

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    Sadly the question has changed. Perhaps a quick edit? – 2017-01-04
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    Done.­­­­­­­­­­ – 2017-01-04