$\underset{x\to 0^{+}}{lim}\frac{x^2\tan x-{\left(\sin x\right)}^3}{{\left(1+x^5\right)}^2-x·\cos x-1+x-\frac{x^3}{2}}$

Sviluppi di Taylor in x=0 :

$\tan x=x+\frac{x^3}{3}+\frac{2}{15}{x}^5+....+o\left(x^{2n+2}\right)$

$\sin x=x-\frac{x^3}{6}+\frac{x^5}{5!}+....+o\left(x^{2m+2}\right)$

$\cos x=1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+....+o\left(x^{2n+1}\right)$

Sviluppo algebrico del numeratore:

$N\left(x\right)=x^2\tan x-{\left(\sin x\right)}^3=x^2\left[x+\frac{x^3}{3}+\frac{2x^5}{15}+...+o\left(x^{2n+2}\right)\right]-{\left[x-\frac{x^3}{6}+\frac{x^5}{5!}+...+o\left(x^{2m+2}\right)\right]}^3=x^3+\frac{x^5}{3}+o\left(x^6\right)-{\left(x-\frac{x^3}{6}+o\left(x^4\right)\right)}^2\left(x-\frac{x^3}{6}+o\left(x^4\right)\right)=$

$=x^3+\frac{x^5}{3}+o\left(x^6\right)-x^3+\frac{x^5}{3}+\frac{x^5}{6}+o\left(x^6\right)=\frac{5}{6}{x}^5+o\left(x^6\right)$

Dall'esame dello sviluppo algebrico si sono dovuti considerare i termini x⁵ perchè i termini x³ si sono semplificati.

Sviluppo algebrico del denominatore (considerando termini fino a x⁵) :

$D\left(x\right)={\left(1+x^5\right)}^2-x·\cos x-1+x-\frac{x^3}{2}=1+x^{10}+2x^5-x·\left(1-\frac{x^2}{2}+\frac{x^4}{24}+o\left(x^5\right)\right)-1+x-\frac{x^3}{2}=$

$=1+x^{10}+2x^5-x+\frac{x^3}{2}-\frac{x^5}{24}+o\left(x^6\right)-1+x-\frac{x^3}{2}=\frac{47}{24}·x^5+o\left(x^6\right)$

In conclusione:

$\underset{x\to 0^{+}}{lim}\frac{x^2\tan x-{\left(\sin x\right)}^3}{{\left(1+x^5\right)}^2-x·\cos x-1+x-\frac{x^3}{2}}=\underset{x\to 0^{+}}{lim}\frac{\frac{5}{6}·x^5+o\left(x^6\right)}{\frac{47}{24}·x^5+o\left(x^6\right)}=\underset{x\to 0^{+}}{lim}\frac{\frac{5}{6}+o\left(x\right)}{\frac{47}{24}+o\left(x\right)}=\frac{5}{6}·\frac{24}{47}=\frac{20}{47}$