I've encountered a rather interesting question which involves finding a limit.

By the instruction to 'find a limit', I can assume that the following limit exists.

$\lim_{x\to 0}$ $\lfloor -x \rfloor +\lfloor 2x+\sqrt2 \rfloor+\lfloor 5x \rfloor$

I have been taught the following, and thus expect that the solution only uses the following concepts:

$\forall$ x ∈ Z, $\lim_{x\to n^-}$ f(x) = n − 1, and $\lim_{x\to n^+}$ f(x) = n.

I have also learned the direct subsititution property, but am not sure whether I can use it as I am not sure whether the limits of the individual terms in the function $\lfloor -x \rfloor +\lfloor 2x+\sqrt2 \rfloor+\lfloor 5x \rfloor$ exist. I suspect they do not as none of the three are continuous as x approaches 0.

I would thus appreciate it if I could be given a push in the right direction. I will gladly offer more information about what I have been taught (and am thus expected to apply).

I believe the answer is 1 - a friend told me - but I would like to understand why it is so.

Thank you.


$\displaystyle\lim_{x \to 0} \lfloor -x\rfloor + \lfloor 2x + \sqrt{2}\rfloor + \lfloor 5x\rfloor = f(x)$ is given to you.

To show that the limit of this as $x$ tends to $0$ is $0$, we have to show that given $\epsilon>0$, there is a $\delta > 0$ such that $|x| < \delta \implies |f(x)| < \epsilon$.

Let us closely examine what is $f(x)$: $$ f(x) = \lfloor -x\rfloor + \lfloor 2x + \sqrt{2}\rfloor + \lfloor 5x\rfloor $$

Note that $|x| < 0.2 \implies \lfloor -x\rfloor + \lfloor 5x\rfloor = -1$. To see why this happens:

1) Suppose that $-0.2 <x < 0$, then $\lfloor -x\rfloor = 0$ and $\lfloor 5x\rfloor = -1$, so their addition is $-1$.

2) Suppose that $0.2 >x > 0$, then $\lfloor -x\rfloor = -1$ and $\lfloor 5x\rfloor = 0$, so their addition is $-1$.

Either way, your expression simplifies to $\lfloor 2x + \sqrt{2}\rfloor - 1$ whenever $|x| < 0.2$.

Now, all we need to do is choose $x$ small enough so that $1 < 2x + \sqrt{2} < 2$, which we get by $\frac{1-\sqrt{2}}{2}<x < \frac{2-\sqrt{2}}{2}$. Thus, whenever $x$ is in this interval (or you can say $x < \min\bigg\{|\frac{1-\sqrt{2}}{2}|,|\frac{2-\sqrt{2}}{2}|\bigg\}$), $\lfloor 2x + \sqrt{2}\rfloor = 1$.

Thus, whenever $|x| < \min \bigg\{0.2,|\frac{1-\sqrt{2}}{2}|,|\frac{2-\sqrt{2}}{2}|\bigg\}$, $f(x) = 0 < \epsilon$.

Hence, the limit is zero.

Now, you cannot substitute values in this question, because this functions is not continuous at zero.

  • $\begingroup$ Thank you very much for this explanation - it took me a few reads to get it, but now I understand that the crux of your method was close observation of the actual function - something I have not trained into myself. However, I admit I have a question. You begin your argument by stating what we are tasked to find, i.e. you begin the argument with stating that the limit is 0. How did you reach this conclusion? Simply through roughly considering $f(0)$? $\endgroup$ – rattat Sep 24 '16 at 11:23
  • $\begingroup$ I thought about it a little bit, and realized why you would come up with such an interval for $x$. Time to apply this thinking to other problems! $\endgroup$ – rattat Sep 24 '16 at 13:44
  • 1
    $\begingroup$ @rattat It is obvious that the floor function is discontinuous at zero. However, we do know one thing, namely that $\lim_{x \to 0^+} f(x) = f(0)$ which is to say it is upper semicontinuous, because the floor function is continuous in a small positive neighbourhood of zero. Because of this property, we can say that the limit is the one approached from above, and while approaching from above I observed the limit is zero. Hence I applied the epsilon-delta method for zero and got the required result. This thinking is very useful if applied to other problems. Good luck in future. $\endgroup$ – астон вілла олоф мэллбэрг Sep 25 '16 at 4:01

A Hint:

Since all the terms in the limit are linear so you can assume that $x\approx 0.1$ where $x\to 0^+$ and then $x\approx -0.1$ where $x\to 0^-$. Now find the value of all terms, one by one, at $0.1$ and $-0.1$ respectively to get the right answer.


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