[tex]\lim_{x \to 0} \frac{\sqrt[3]{1+x} - \sqrt[3]{1-x} }{\sqrt[2]{1+x} - \sqrt[2]{1-x} }[/tex] Решите пожалуйста лимит. Created by lepme. algebra-ru

tex] Решите пожалуйста лимит...

армения20171

$\lim\limits _{x \to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\\\\=\lim\limits_{x \to 0}\frac{(\sqrt[3]{1+x}-\sqrt[3]{1-x})(\sqrt[3]{(1+x)^2}+\sqrt[3]{(1+x)(1-x)}+\sqrt[3]{(1-x)^2})(\sqrt{1+x}+\sqrt{1-x})}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})(\sqrt[3]{(1+x)^2}+\sqrt[3]{(1+x)(1-x)}+\sqrt[3]{(1-x)^2})}=\\\\=\lim\limits_{x \to 0}\frac{(1+x-1+x)(\sqrt{1+x}+\sqrt{1-x})}{(1+x-1+x)(\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2})}=\frac{1+1}{1+1+1}=\frac{2}{3}$

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Verified answer

Lim(³√(1+x)-³√(1-x))/(√(1+x)-√(1-x))
x->0

=lim((1+x)^1/6)²-(1-x)^1/6)²)/((1+x)^1/6)³-(1-x)^1/6)³)=

lim(1+x)^1/6+(1-x)^1/6)/

((1+x)^1/6)²+((1+x)(1-x))^1/6+
((1-x)^1/6)²)=(1+1)/(1+1+1)

=2/3

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