Calcolare, in funzione di α, β ∈ ℝ\{0} il limite indicato: $\underset{x\to 0}{lim}\frac{x^3+x^4}{[{\left(1+x^2\right)}^{\beta }-1]\tan \left(\alpha x\right)}$

Sviluppi di Taylor:

${\left(1+x^2\right)}^{\beta }-1={\beta x}^2+o\left(x^2\right)$

$\tan \left(\alpha x\right)=\alpha x+o\left({\alpha }^2{x}^2\right)$

Sviluppo del denominatore:

$N\left(x\right)=[{\left(1+x^2\right)}^{\beta }-1]\tan \left(\alpha x\right)=(\beta x^2+o\left(x^2\right))·(\alpha x+o\left({\alpha }^2{x}^2\right))=\mathrm{\alpha \beta }{x}^3+o\left(\alpha x^4\right)$

$\underset{x\to 0}{lim}\frac{x^3+x^4}{[{\left(1+x^2\right)}^{\beta }-1]\tan \left(\alpha x\right)}=\underset{x\to 0}{lim}\frac{x^3+x^4}{\mathrm{\alpha \beta }{x}^3+o\left(\alpha x^3\right)}=\underset{x\to 0}{lim}\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\alpha $}\right.+\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$\alpha $}\right.}{\beta +o\left(1\right)}=\frac{1}{\alpha \beta }$