[m]ln(1+∛x) ∼ ∛x[/m]
[m]sin∛x ∼ ∛x[/m] ⇒ [m]sin^4x=sinx\cdot sinx\cdot sinx\cdot sinx ∼ ∛x\cdot ∛x\cdot ∛x\cdot ∛x=x\cdot ∛x[/m]
и
[m] lim_{x → 0}(1+x)^{\frac{1}{x}}=e[/m]
[m] lim_{x → 0}\frac{∛x\cdot x}{sin^4∛x}= lim_{x → 0}\frac{∛x\cdot x}{∛x\cdot x}=1[/m]
Решение:
[m]lim_{x → 0}(1-ln(1+∛x))^{\frac{x}{sin^4∛x}}=lim_{x → 0}((1-∛x)^{(-\frac{1}{∛x})})^{(-\frac{∛x\cdot x}{sin^4∛x})}=e^{lim_{x →0}(-\frac{∛x\cdot x}{sin^4∛x})}=[/m][m]e^{-1}=\frac{1}{e} [/m]