Question

$\underset{x\to \mathrm{\pi }}{\lim }\frac{1+\cos x}{{\tan }^{2}x}$

Answer 1

$$\begin{gathered} \underset{x\to \mathrm{\pi }}{\lim }\frac{1+\cos x}{{\tan }^{2}x} \\ =\underset{h\to 0}{\lim }\frac{1+\cos \left(\mathrm{\pi }+h\right)}{{\tan }^2\left(\mathrm{\pi }+h\right)} \\ =\underset{h\to 0}{\lim }\frac{1-\cos h}{{\tan }^{2}h} \\ =\underset{h\to 0}{\lim }\frac{2{\sin }^2{\displaystyle \frac{h}{2}}}{{\tan }^{2}h} \\ =\underset{h\to 0}{\lim }\frac{2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle 4\left(\frac{h^2}{4}\right)\times \frac{{\tan }^{2}h}{h^2}}} \\ =\frac{1}{2}\frac{\underset{h\to 0}{\lim }{\left({\displaystyle \frac{\sin {\displaystyle \frac{h}{2}}}{\frac{h}{2}}}\right)}^2}{{{\displaystyle \underset{h\to 0}{\lim }\left(\frac{\tan h}{h}\right)}}^2} \\ =\frac{1}{2}\times \frac{{\left(1\right)}^2}{{\left(1\right)}^2}=\frac{1}{2} \end{gathered}$$