Fórum de Matemática
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Olá, tentei resolver com diferença de cubos mas não consegui



\(\lim \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}\)
X--->0

\(\\\\\\ \lim_{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}\)



\(\\\\\\ \lim_{x \rightarrow 0} \frac{(\sqrt{1+x}-\sqrt{1-x})*(\sqrt{1+x}+\sqrt{1-x})*(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}*\sqrt[3]{1-x}+\sqrt[3]{(1-x)^{2}})}{(\sqrt[3]{1+x}-\sqrt[3]{1-x})*(\sqrt{1+x}+\sqrt{1-x})*(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}*\sqrt[3]{1-x}+\sqrt[3]{(1-x)^{2}})}\)


\(\\\\\\ \lim_{x \rightarrow 0} \frac{(1+x-(1-x))*(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}*\sqrt[3]{1-x}+\sqrt[3]{(1-x)^{2}})}{(1+x-(1-x))*(\sqrt{1+x}+\sqrt{1-x})}\)


\(\\\\\\ \lim_{x \rightarrow 0} \frac{(1+x-1+x)*(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}*\sqrt[3]{1-x}+\sqrt[3]{(1-x)^{2}})}{(1+x-1+x)*(\sqrt{1+x}+\sqrt{1-x})}\)



\(\\\\\\ \lim_{x \rightarrow 0} \frac{2x*(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}*\sqrt[3]{1-x}+\sqrt[3]{(1-x)^{2}})}{2x*(\sqrt{1+x}+\sqrt{1-x})}\)



\(\\\\\\ \lim_{x \rightarrow 0} \frac{\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}*\sqrt[3]{1-x}+\sqrt[3]{(1-x)^{2}}}{\sqrt{1+x}+\sqrt{1-x}}\)


veja que não possui mais indeterminação,já podemos substituir:


\(\\\\\\ \frac{\sqrt[3]{(1+0)^{2}}+\sqrt[3]{1+0}*\sqrt[3]{1-0}+\sqrt[3]{(1-0)^{2}}}{\sqrt{1+0}+\sqrt{1-0}}\)


\(\\\\\\ \frac{3}{2}\)