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Question Number 186983 by cortano12 last updated on 12/Feb/23

Answered by CElcedricjunior last updated on 12/Feb/23

$$\underset{x\to 0}{\lim }\frac{\mathit{x}\sqrt[3]{\mathit{x}-1}+\sqrt[4]{(\mathit{x}+1)}-1}{{\mathit{x}}^2\sqrt[3]{\mathit{x}+1}+\sqrt[4]{\mathit{x}+1}-1}=\mathit{k}$$$$\mathit{k}=\underset{x\to 0}{\lim }\frac{\mathit{x}(-1-\frac{1}{3}\mathit{x}+\frac{1}{9}{\mathit{x}}^2)+(1+\frac{1}{4}\mathit{x}-\frac{3}{32}{\mathit{x}}^2)-1}{{\mathit{x}}^2(1+\frac{1}{3}\mathit{x}-\frac{1}{9}{\mathit{x}}^2)+(1+\frac{1}{4}\mathit{x}-\frac{3}{32}{\mathit{x}}^2)-1}★celebrecedricjunior$$$$=\underset{x\to 0}{\lim }\frac{-\mathit{x}-\frac{1}{3}{\mathit{x}}^2+\frac{1}{9}{\mathit{x}}^3+\frac{1}{4}\mathit{x}-\frac{3}{32}{\mathit{x}}^2}{{\mathit{x}}^2+\frac{1}{3}{\mathit{x}}^3-\frac{1}{9}{\mathit{x}}^4+\frac{1}{4}\mathit{x}-\frac{3}{32}}$$$$=\frac{\frac{1}{4}-1}{\frac{1}{4}}=-3$$$$\underset{x\to 0}{\lim }\frac{\mathit{x}\sqrt[3]{\mathit{x}-1}+\sqrt[4]{\mathit{x}+1}-1}{{\mathit{x}}^2\sqrt[3]{\mathit{x}+1}+\sqrt[4]{\mathit{x}+1}-1}=-3★Moivre$$

Answered by ARUNG_Brandon_MBU last updated on 12/Feb/23

$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{x\sqrt[3]{x-1}+\sqrt[4]{x+1}-1}{x^2\sqrt[3]{x+1}+\sqrt[4]{x+1}-1}$$$$=\underset{x\to 0}{\lim }\frac{-x(1-\frac{x}{3})+(1+\frac{x}{4})-1}{x^2(1+\frac{x}{3})+(1+\frac{x}{4})-1}$$$$=\underset{x\to 0}{\lim }\frac{\frac{x^2}{3}-\frac{3x}{4}}{\frac{x^3}{3}+x^2+\frac{x}{4}}=\underset{x\to 0}{\lim }\frac{\frac{x}{3}-\frac{3}{4}}{\frac{x^2}{3}+x+\frac{1}{4}}=-3$$

Answered by cortano12 last updated on 14/Feb/23

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