Question

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)\sin x-2\cos x}{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)+\cot x} \\ =\underset{h\to 0}{\lim }\frac{\left[{\displaystyle \frac{\mathrm{\pi }}{2}}-\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)\right]\sin \left(\frac{\mathrm{\pi }}{2}-h\right)-2\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}{\left[{\displaystyle \frac{\mathrm{\pi }}{2}}-\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)\right]+\cot \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}\left(\mathrm{Plugging}x=\frac{\mathrm{\pi }}{2}-h\right) \\ =\underset{h\to 0}{\lim }\frac{h\cos h-2\sin h}{h+\tan h} \\ \text{Dividing the numerator and the denominator by h:} \\ \underset{h\to 0}{\lim }\frac{\cos h-2{\displaystyle \frac{\sin h}{h}}}{1+{\displaystyle \frac{\tan h}{h}}} \\ \Rightarrow \frac{1-2\left(1\right)}{1+1}=\frac{-1}{2} \end{gathered}$$