$$\underset{x\to +\mathrm{\infty }}{lim}\frac{{\left(1+x^2\right)}^{-\frac{1}{1+{\lambda }^2}}}{{\left(\frac{1}{x}\right)}^n}=\underset{x\to +\mathrm{\infty }}{lim}\frac{x^n}{{\left(1+x^2\right)}^{\frac{1}{1+{\lambda }^2}}}=\underset{x\to +\mathrm{\infty }}{lim}{\left[\frac{x^{n(1+{\lambda }^2)}}{1+x^2}\right]}^{\frac{1}{1+{\lambda }^2}}$$