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