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Question Number 129120 by liberty last updated on 13/Jan/21

$$\underset{(x,\mathrm{y})\to (\mathrm{\infty },\mathrm{\infty })}{\lim }{(\frac{\pi }{2}-\arctan ({\mathrm{x}}^2+{\mathrm{y}}^2))}^{\frac{1}{\ln ({\mathrm{x}}^2+{\mathrm{y}}^2)}}=?$$

Answered by benjo_mathlover last updated on 13/Jan/21

$$\mathcal{L}=\underset{(x,\mathrm{y})\to (\mathrm{\infty },\mathrm{\infty })}{\lim }{(\frac{\pi }{2}-{\tan }^{-1}({\mathrm{x}}^2+{\mathrm{y}}^2))}^{\frac{1}{\ln ({\mathrm{x}}^2+{\mathrm{y}}^2)}}$$$$\mathcal{L}={\mathrm{e}}^{\underset{(x,\mathrm{y})\to (\mathrm{\infty },\mathrm{\infty })}{\lim }\left(\frac{\frac{\pi }{2}-{\tan }^{-1}({\mathrm{x}}^2+{\mathrm{y}}^2)}{\ln ({\mathrm{x}}^2+{\mathrm{y}}^2)}\right)}$$$$\mathcal{L}={\mathrm{e}}^{\underset{(x,\mathrm{y})\to (\mathrm{\infty },\mathrm{\infty })}{\lim }\left(\frac{-\frac{2\mathrm{x}+2\mathrm{y}}{1+{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}}{\frac{2\mathrm{x}+2\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}}\right)}$$$$\mathcal{L}={\mathrm{e}}^{\underset{(x,\mathrm{y})\to (\mathrm{\infty },\mathrm{\infty })}{\lim }-\left(\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{1+{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}\right)}$$$$\frac{\mathcal{L}={\mathrm{e}}^0=1}{}$$

Answered by mathmax by abdo last updated on 14/Jan/21

$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{rcos}\theta \mathrm{and}\mathrm{y}=\mathrm{rsin}\theta \Rightarrow$$$$\mathrm{Lim}={\lim }_{\mathrm{r}\to +\mathrm{\infty }}{(\frac{\pi }{2}-\arctan \left({\mathrm{r}}^2\right))}^{\frac{1}{2\mathrm{lnr}}}={\lim }_{\mathrm{r}\to \mathrm{\infty }}{\mathrm{e}}^{\frac{1}{2\mathrm{lnr}}\ln (\frac{\pi }{2}-\arctan \left({\mathrm{r}}^2\right))}$$$$\mathrm{we}\mathrm{have}\mid \frac{\pi }{2}-\arctan \left({\mathrm{r}}^2\right)\mid <\pi \Rightarrow \frac{1}{2\mathrm{lnr}}\ln (\frac{\pi }{2}-\arctan \left({\mathrm{r}}^2\right))<\frac{\pi }{2\mathrm{lnr}}\to \mathrm{o}(\mathrm{r}\to +\mathrm{\infty })\Rightarrow$$$${\lim }_{\mathrm{r}\to 0}{\mathrm{e}}^{\frac{1}{2\mathrm{nr}}\ln (\frac{\pi }{2}-\arctan \left({\mathrm{r}}^2\right))}={\mathrm{e}}^0=1$$

Commented by mathmax by abdo last updated on 14/Jan/21

$$0<\theta <\frac{\pi }{2}$$

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