# Consider the sequence ${x_n}$ defined by $x_n = [nx]/ n$

Consider the sequence ${x_n}$ defined by $x_n = [nx]/ n$ for $x\in\mathbb R$ where $[·]$ denotes the integer part. Then ${x_n}$
(a) converges to $x$.
(b) converges but not to $x$.
(c) does not converge
(d) oscillates.

I think (a) is correct as $\lim_{n \to \infty}[n]/n=1$. Am I right?

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answer should depend on x. – i. m. soloveichik Nov 23 '12 at 14:20

You are right since for $x=1$ you get $\displaystyle{\lim_{n \to \infty}\frac{[xn]}{n}=\lim_{n \to \infty}\frac{[n]}{n}=1=x}$.
Therefore you can eliminate (b),(c) and (d) and the correct answer is (a).
For the proof use that $$nx-1<[nx]\leq nx\Rightarrow x-\frac{1}{n}<\frac{[nx]}{n}\leq x, \ \forall n \in \mathbb{N}.$$

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We have $|\,\lfloor nx\rfloor -nx\,|<1$, which implies $$\left|\frac{\lfloor nx\rfloor}n - x\right|<\frac1n.$$ So $x_n$ always converges to $x$, for any $x\in\mathbb R$.

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The answer is (a), but your reasoning is not correct: you cannot factor out $x$ like that. What you can do is observe that

$$0\le x-\frac{\lfloor nx\rfloor}n=\frac{nx-\lfloor nx\rfloor}n<\frac1n$$

and take limits as $n\to\infty$.

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