Evaluating the Limit of a Sum of Floor Functions

Question

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.

Answer 1

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

Answer 2

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.