Floor of $\cos x$

As $x\rightarrow 0, \text{cos }x \rightarrow 1$ but always less than $1$ . So, $\lfloor \text{cos }x\rfloor$ = 0. Now we have

$\displaystyle \lim_{x\to 0} \text{cot}^{-1} 0 = \frac{\pi}{2} \approx 1.57$

Edit : it depends on your definition of limits.

The right answer is : the limit does not exist

Proof : Let $f(x)=\cot^{-1} ( \lfloor \cos(x) \rfloor)$ .

Let $x_n = \dfrac{1}{n}$ and $y_n=0$ . $x_n,\ y_n \underset{n\to+\infty}{\longrightarrow} 0$

Therefore if the limit exists (call it $L$ ), then we should have $f(x_n), \ f(y_n) \underset{n\to+\infty}{\longrightarrow} L$

But $f(x_n)=\dfrac{\pi}{2}$ and $f(y_n)=\dfrac{\pi}{4}$ so the limit does not exist.