# For which $a$ does $\lim_{x\to a}f(x) = \lfloor x \rfloor$ exist?

### How do I go about finding for which values of $a$ the $\lim_{x\to a}f(x) = \lfloor x \rfloor$ doesn't exist?

I've tried using $x-1 \lt \lfloor x \rfloor \le x$, but it seems I'm lacking some fundamental way of thinking about the general picture... I'm a newbie in the subject, so a meticulous answer would be much appreciated... :)

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You should draw the graph of the floor function. There is no trick. You can read the answer on the graph. – 1015 Mar 26 '13 at 22:07
I should add: once you know the answer, it is easier to prove it formally. – 1015 Mar 26 '13 at 22:15
Hi, @julien, could you correct my following reasoning? Suppose $a$ is integer so $a = \lfloor a \rfloor$: If $\lim_{x\to a^-} \lfloor x\rfloor = \lfloor a\rfloor = a$, then $\varepsilon - a \lt \lfloor x \rfloor \lt \varepsilon + a.$ But $x - 1 < \lfloor x \rfloor \lt \varepsilon + a$ and therefore, if the limit exists at $a$, it must satisfy "$x - a \lt \delta \to x - a < \varepsilon + 1$" which is true $\forall x$ if $\delta \lt \varepsilon + 1$. – FRD Mar 26 '13 at 22:25
Hi, @FRD. Your approach is not very natural. If $a$ is an integer, prove that the limit does not exist. If $a$ is not an integer, prove that the limit exists and is what you want. Just like Brian M. Scott did. You should not try to go by contradiction in such a situation. – 1015 Mar 26 '13 at 23:42
Contrived...I've learned a new word, thanks. But not, this is far from contrived. Like I said in my comments above: draw the graph of the floor function first. This distinction is then more than natural. Well, from the definition of the function too, but in a less visual way. – 1015 Mar 26 '13 at 23:51

Suppose that $a$ is not an integer, and let $n=\lfloor a\rfloor$; then $n<a<n+1$, and the floor function is constant on the interval $(n,n+1)$. For all $x$ sufficiently close to $a$, $x$ is in the interval $(n,n+1)$, and $\lfloor x\rfloor=n=\lfloor a\rfloor$. Thus,

$$\lim_{x\to a}\lfloor x\rfloor=n=\lfloor a\rfloor\;,$$

and $f(x)=\lfloor x\rfloor$ is continuous at $a$.

To put it a little differently, on the open interval $(n,n+1)$ the functions $f(x)=\lfloor x\rfloor$ and $g(x)=n$ are completely indistinguishable, and $g$ is continuous, so $f$ is also continuous on this interval. (It’s important here that the interval is open.)

Now suppose that $a$ is an integer; then $f(a)=\lfloor a\rfloor=a$. But for every $x<a$ we have $f(x)\le a-1$, so

$$\lim_{x\to a^-}f(x)\ne a=f(a)\;,$$

and $f$ is not continuous from the left at $a$. Of course this means that $f$ is not continuous at $a$, though you should try to convince yourself that $f$ is continuous from the right at $a$.

Added: Note that it’s not clear a priori that $\lim\limits_{x\to a^-}f(x)$ even exists. Of course if it doesn’t exist, then it certainly isn’t equal to $a$. However, it does exist, and it would be a nice little exercise to verify that it’s equal to $a-1$.

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How can I find "for every x<a we have f(x)≤a−1" from the equations only, without looking at the graph? – FRD Mar 26 '13 at 22:39
@FRD: It’s always true that $\lfloor x\rfloor\le x$, so if $x<a$, we have $\lfloor x\rfloor\le x<a$. Since $a$ is an integer, and $\lfloor x\rfloor$ is an integer less than $a$, the largest possible value of $\lfloor x\rfloor$ is $a-1$. – Brian M. Scott Mar 26 '13 at 22:42
Detail: given what you've said in the second step of your argument, what you can conclude is $\limsup_{a^-}f\leq a-1\neq a=f(a)$. Of course, the left limit exists and is equal to $a-1$. But you have only mentioned the inequality $f(x)\leq a-1$. – 1015 Mar 26 '13 at 23:46
@julien: I can conlude exactly what I did conclude: that $\lim_{x\to a^-}f(x)\ne a$. I don’t care whether equality fails because the limit doesn’t exist or because it’s something other than $a$ (as is of course the case); all that matters is it cannot be $a$. – Brian M. Scott Mar 26 '13 at 23:49
I understood that. It's just that when I read: $\lim_{a^-} f$ does or does not do something, I understand that this limit exists. If it does not exist, I don't use the symbol. Maybe I'm in a minority. But you're the one who also told me one should write $f[X]$ and not $f(X)$ to denote the range of a function. So I thought you would have agreed with me on this. Not a big deal anyway of course. – 1015 Mar 26 '13 at 23:56

Hint: Look at limits $\lim_{x\to a^+}f(x)$ and $\lim_{x\to a^-}f(x)$ for any whole number $a$ (try for example $a=1$).

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I was looking for an answer involving the epsilon-delta notation, and I should have explicitly said so. I think I've found it, so here it is:

Starting off, we have $\lfloor a \rfloor - \varepsilon < \lfloor x \rfloor < \lfloor a\rfloor+\varepsilon$.
It follows that $\lfloor x\rfloor$ is

• at least: the closest integer value greater than $\lfloor a\rfloor - \varepsilon$, which is $\lceil \lfloor a\rfloor - \varepsilon\rceil$. Then $$\lceil \lfloor a\rfloor - \varepsilon\rceil = \lfloor a\rfloor - \lfloor \varepsilon\rfloor \le \lfloor x\rfloor \le x$$
• at most: the closest integer value smaller than $\lfloor a\rfloor + \varepsilon$, which is $\lfloor \lfloor a\rfloor + \varepsilon\rfloor$. Then $$\lfloor \lfloor a\rfloor + \varepsilon\rfloor = \lfloor a\rfloor + \lfloor \varepsilon\rfloor \ge \lfloor x\rfloor$$

Now, to prove that the limit exists, we must show that $$\forall \ \varepsilon > 0, \ \exists \ \delta>0 \mathtt{\ such\ that, if} \ \delta > |a-x| >0$$ $$\Rightarrow \lfloor a\rfloor + \lfloor \varepsilon\rfloor > \lfloor x\rfloor > \lfloor a \rfloor + \lfloor \varepsilon \rfloor$$

Take $\varepsilon = 0.5$.
The equation above yields: $$\lfloor a \rfloor \ge x \ge \lfloor a \rfloor \Rightarrow \lfloor x \rfloor = \lfloor a \rfloor \le min(x, a)$$

If $\delta > |a-x| >0$, it might follow that $x<a$ which can only hold true if $\lfloor a \rfloor \lt a$ (that is, if $a$ isn't an integer) since then we can have $\lfloor x \rfloor = \lfloor a \rfloor \lt x \lt a$.

Therefore for any $\delta>0$ and $a$ such that $\delta > |a - \lfloor a \rfloor| > |a - x| >0$, the condition for the limit will hold true.

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