I recently started studying continuity and I tried to figure out continuity of $\lfloor \arctan x\rfloor$ using graphs. I got a similar graph to the one which is in the Desmos graph plotter.

To check the graph of $\lfloor \arctan x\rfloor$ in Desmos graph plotter visit the link given below:


My doubt is that at $x=\tan(1)=1.557$ the value of $\lfloor \arctan x\rfloor$ has to be one, but in the graph of Desmos website the value given is $0$.

How is that possible? Could anyone explain the graph and solve my doubt? Any little help is appreciable.

Moreover, you can check and share about the Desmos website, it is a very good website and can plot even complex graphs.

  • $\begingroup$ 1 in radians right? $\endgroup$
    – Chris2018
    Jun 24, 2018 at 4:39
  • $\begingroup$ what do you mean i graphed y=tan(x) on desmos at x=1 the the y coordinate was about 1.557 $\endgroup$
    – Chris2018
    Jun 24, 2018 at 4:42
  • $\begingroup$ Just add the function invtanx in desmos plotter.It shows that point (1.557,0) lies on x axis while I think it should be (1.557,1).Yes,the '1' is in radians $\endgroup$
    – Banchin
    Jun 24, 2018 at 4:52

2 Answers 2


$$\tan(1)\approx 1.557407724654902230506974807458360173087250772381520038>1.557\implies \arctan 1.557<1$$(See http://m.wolframalpha.com/input/?i=tan%281%29 )

Note that $\arctan 1.557$ is very close to $1$, but the floor function is not continuous hence unless it is greater or equal how close it is doesn't matter: $\arctan 1.557\approx 0.99981$( see http://m.wolframalpha.com/input/?i=arctan%281.557%29 )


We have $$ \begin{cases} -{\large{\frac{\pi}{2}}} < \arctan(x) < -1&\;\;\;\text{if}\;\,x < -\tan(1)\\[4pt] -1\le \arctan(x) < 0&\;\;\;\text{if}\;\,-\tan(1) \le x < 0\\[4pt] 0\le \arctan(x) < 1&\;\;\;\text{if}\;\,0 \le x < \tan(1)\\[4pt] 1 \le \arctan(x) < {\large{\frac{\pi}{2}}} &\;\;\;\text{if}\;\,x \ge \tan(1)\\[4pt] \end{cases} $$ Noting that $1 < {\large{\frac{\pi}{2}}} < 2$, we get $$ \lfloor{\arctan(x)}\rfloor= \begin{cases} -2&\;\;\;\text{if}\;\,x < -\tan(1)\\[4pt] -1&\;\;\;\text{if}\;\,-\tan(1) \le x < 0\\[4pt] 0&\;\;\;\text{if}\;\,0 \le x < \tan(1)\\[4pt] 1&\;\;\;\text{if}\;\,x \ge \tan(1)\\[4pt] \end{cases} $$ Thus, the graph of the function $$f(x)=\lfloor{\arctan(x)}\rfloor$$ is piecewise constant, with $4$ horizontal pieces, one at each of the $y$-values $-2,-1,0,1$.

Thus, based on the piecewise constant form of $f$, we see that if $x=\tan(1)$, then $f(x)=1$, but if $x$ is slightly less than $\tan(1)$, then $f(x)=0$.

Hence, when evaluating $f(x)$ for decimal values of $x$ near $\tan(1)$, you need to worry about the possibility that $x$ might be slightly less than $\tan(1)$.

In particular, since $1.557$ is slightly less than $\tan(1)$, we get $f(1.557)=0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .