Solución 1
Usar la fórmula McLaughlin de Hinks para expandir
sin6x=6x-(6x) ^ 3/3! o(x^3)
F(x) expande f(x)= f(0) f '(0)x 1/2f ' '(0)x2 o(x2)
Obtener en su lugar
lim[sin6x xf(x)]/x^3=6x-(6x)^3/3! o(x^3) f(0)x f'(0)x^2 1/2f''(0)x^3 o(x^3)/x^3=0 x→0
lim[6x f(0)x f '(0)x 2]/x 3 1/2f ''(0)-36 = 0.
Entonces f(0)=-6 f '(0)= 0 1/2 f ' '(0)-36 = 0 f ' '(0)= 72.
Lin[6 f(x)]/x^2=limf''(0)/2=36
Solución 2
lim[6 f (x)]/x^2=lim[6 f(x)]/x^2-0
=lim[6 f(x)]/x^2-lim[sin6x xf( x )]/x^3
=lim{[6 f(x)]/x^2-[sin6x xf(x)]/x^3}
= lim [ (6x-sin6x)/x 3](ley de Lópida)
=lim[(6-6cos6x)/3x^2]
= 2lim [(1- cos6x)/ x 2](sustitución infinitesimal equivalente)
=2lim[(1/2)(6x)^2/x^2]
=6^2= 36