$\lim_{x->0}\frac{\sqrt{x+1}+\sqrt[3]{x-1}}{x}$
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Posted 06-01-2015 - 23:08
vandong98
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$\lim_{x\rightarrow 0}\frac{\sqrt{x+1}+\sqrt[3]{x-1}}{x}=\lim_{x\rightarrow 0}\frac{(\sqrt{x+1}-1)+(\sqrt[3]{x-1}+1)}{x}=\lim_{x\rightarrow 0}{\frac{\frac{x}{\sqrt{x+1}+1}+\frac{x}{\sqrt[3]{x-1}^{2}-\sqrt[3]{x-1}+1}}{x}}=\lim_{x\rightarrow 0}(\frac{1}{\sqrt{x+1}+1}+\frac{1}{\sqrt[3]{x-1}^{2}-\sqrt[3]{x-1}+1})=\frac{3}{2}$
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