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Here is my question. Using the definition of limits, prove that the function $f(x)= \left \lfloor x \right \rfloor$ is discontinuous at every integer.

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    That's not a question; it's a command. It looks like it might be a verbatim quote of a homework exercise, in which case: What have you tried? Do you know the definition of $\lfloor x\rfloor$? Do you know the definition of (dis)continuous? Limits?2012-09-21

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Hint: For each integer $n$, take one sequence converging to $n$ from above, and another converging to $n$ from below.

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    Oh, very true. I was blinded by the (my?) standard way of solving these problems. Not sure I can think of a good hint for doing it directly from the usual limit of functions definition, but maybe somebody else can.2012-09-21
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The first two paragraphs below are background, necessary for understanding what's going on, but not part of the formal argument. The last two paragraphs give the actual proof.

Let's find out what's happening. Take your favourite integer, maybe $13$. Is $f$ continuous at $13$? As you travel in a rightward direction from $12$, $f(x)$ is for a while steadily $12$, and then jumps suddenly at $13$, since $f(13)=13$. This jump is what kills continuity at $13$. If $x$ is close to $13$ but below $13$, then $f(x)$ is nowhere near $f(13)$.

A bit more formally, is it true that for any $\epsilon\gt 0$, then as long as $x$ is close enough to $13$, then $f(x)$ will be within $\epsilon$ of $13$? Can we find a $\delta$ such that if $|x-13|\lt \delta$, then $|f(x)-f(13)|\lt \epsilon$? Take for example $\epsilon=1/10$. Is it true that there is a $\delta$ such that if $|x-13|\lt \delta$, then $|f(x)-13|$ is guaranteed to be $\lt 1/10$? No, because no matter how close $x$ is to $13$, as long as $x\lt 13$, $f(x)$ will be "far" from $13$. More precisely, we will have $|f(x)-13|\ge 1\gt 1/10$. Now we can write the proof. It is short.

Let $a$ be an integer. Then $f(a)=a$. Let $\epsilon=1/10$. We show that there is no $\delta\gt 0$ such that if $|x-a|\lt \delta$, then $|f(x)-f(a)|\lt \epsilon$.

For suppose that $\delta\gt 0$, and let $x=a-\delta/2$. Then $f(x)\lt a-1$, and therefore $|f(x)-f(a)|\ge 1\gt \epsilon$.

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    You say they probably have an intuitive feeling, but we don't really know that. It sometimes happens that someone will post a question like this and then reveal in the comments that what they need is for the question to be explained, much more than the answer given. (I don't mean to go on, this is a good answer, just trying to explain my concern).2012-09-21