[tex]l=\underset{x\rightarrow \infty}{lim} x(\pi-2arctgx) $=$ \underset{x\rightarrow \infty}{lim} \dfrac{\pi-2arctgx}{\dfrac{1}{x}} \overset{(L'H.)\frac{0}{0}}{=}$\underset{x\rightarrow \infty}{lim} \dfrac{(\pi-2arctgx)'}{\Big(\dfrac{1}{x}\Big)'}= \\ \\ =\underset{x\rightarrow \infty}{lim}$ $ \dfrac{0- \dfrac{2}{x^2+1} }{-\dfrac{1}{x^2}} = \underset{x\rightarrow \infty}{lim}$ $\dfrac{2}{x^2+1}\cdot \dfrac{x^2}{1}= \underset{x\rightarrow \infty}{lim}$ $\dfrac{2x^2}{x^2+1} = 2[/tex]
[tex]$ \ $Raspuns corect: $(a).[/tex]
