# $\lim_{x \rightarrow \pi}\left \lfloor \frac{3\sin(x^2)-1}{2\cos(x)+1} \right \rfloor$=?

How to find this limit: $$\lim_{x \rightarrow \pi}\left \lfloor \frac{3\sin(x^2)-1}{2\cos(x)+1} \right \rfloor$$ where $\left \lfloor x \right \rfloor$ is the floor-function?

Generally, how to find limit of floor-function? tnx

• Could you find the limit if the floor wasn't there? – Henrik Nov 17 '16 at 14:26

Hint. The floor function $t\to\lfloor t\rfloor$ is continuous for $t\in \mathbb{R}\setminus \mathbb{Z}$. On the other hand for $n\in\mathbb{Z}$, $$\lim_{t \rightarrow n^+}\lfloor t\rfloor=n\quad\mbox{and}\quad \lim_{t \rightarrow n^-}\lfloor t\rfloor=n-1.$$ Evaluate $\displaystyle \lim_{x \rightarrow \pi}\frac{3\sin(x^2)-1}{2\cos(x)+1}.$ Is it an integer number?