Let $\left[\mathrm{x}\right]$ denote the greatest integer not exceeding $\mathrm{x}.$ If 
${\mathcal{l}}_1=\underset{\mathrm{x}\to 2^{+}}{\lim }\left({\mathrm{x}}^2+\left[\mathrm{x}\right]\right), {\mathcal{l}}_2=\underset{\mathrm{x}\to 3^{-}}{\lim }\left(2\mathrm{x}-\left[\mathrm{x}\right]\right)$ and ${\mathcal{l}}_3=\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\lim }\left(\frac{\mathrm{cosx}}{\mathrm{x}-{\displaystyle \frac{\mathrm{\pi }}{2}}}\right),$ then

$\begin{array}{rcl}{\mathrm{L}}_1& =& \underset{\mathrm{x}\to 2+}{\mathrm{Lt}}{\mathrm{x}}^2+\left[\mathrm{x}\right]\\ & =& \underset{\mathrm{x}\to 2+}{\mathrm{Lt}} {\mathrm{x}}^2+2\\ & =& 4+2=6\end{array}$

$$\begin{gathered} {\mathrm{L}}_2=\underset{\mathrm{x}\to 3-}{\mathrm{Lt}} 2\mathrm{x}-\left[\mathrm{x}\right] \\ =\underset{\mathrm{x}\to 3-}{\mathrm{Lt}} 2\mathrm{x}-2 \\ =6-2=4 \end{gathered}$$

$$\begin{gathered} {\mathrm{L}}_3=\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{Lt}} \frac{\cos \mathrm{x}}{\mathrm{x}-{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{Lt}}\frac{\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)}{-\left(\frac{\mathrm{\pi }}{2}-\mathrm{x}\right)} \\ =-1 \\ {\mathrm{L}}_3<{\mathrm{L}}_2<{\mathrm{L}}_1 \end{gathered}$$