Floor of $\cos x$ Part 2

Note that

$\begin{aligned} L^+ & = \lim_{x \to \frac \pi 2^+} \tan^{-1} \lfloor \cos (x) \rfloor = \tan^{-1} 0 = 0 \\ L^- & = \lim_{x \to \frac \pi 2^-} \tan^{-1} \lfloor \cos (x) \rfloor = \tan^{-1} -1 = - \frac \pi 4 \end{aligned}$

Since $L^+ \ne L^-$ , limit $L = \displaystyle \lim_{x \to \frac \pi 2} \tan^{-1} \lfloor \cos (x) \rfloor$ does not exist .

The function $\lfloor \cos x \rfloor$ has a jump discontinuity at $x = \frac{\pi}{2}$ . This function equals $0$ for $0 < x \leq \frac{\pi}{2}$ , but equals $-1$ for $\frac{\pi}{2} < x < \frac{3\pi}{2}$ .

And since the inner function has a jump discontinuity, so does the outer function. Because $\tan^{-1}(0) = 0$ while $\tan^{-1} (-1)= \frac{-\pi}{4}$ .