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Question Number 119977 by bramlexs22 last updated on 28/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\left(\sqrt[5]{x^4}(\sqrt[5]{x+1}-\sqrt[5]{x})\right)=?$$

Answered by bemath last updated on 28/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[5]{x^5+x^4}-\sqrt[5]{x^5}=\underset{x\to \mathrm{\infty }}{\lim }x\sqrt[5]{1+\frac{1}{x}}-x$$$$[let\frac{1}{x}=v]$$$$\underset{v\to 0}{\lim }\frac{\sqrt[5]{1+v}-1}{v}=\underset{v\to 0}{\lim }\frac{(1+\frac{v}{5})-1}{v}=\frac{1}{5}$$

Answered by Dwaipayan Shikari last updated on 28/Oct/20

$$\sqrt[5]{x^4}(\sqrt[5]{x}(\sqrt[5]{1+\frac{1}{x}}-1)$$$$\sqrt[5]{x^5}(1+\frac{1}{5x}-1)$$$$=x.\frac{1}{5x}=\frac{1}{5}$$

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