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Question Number 174258 by blackmamba last updated on 28/Jul/22

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{\sqrt{x^4-x^3}}-x=?$$

Answered by cortano1 last updated on 28/Jul/22

$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x^4-x^3}-x^2}{\sqrt{\sqrt{x^4-x^3}}+x}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^2(\sqrt{1-\frac{1}{x}}-1)}{\sqrt{x^2\sqrt{1-\frac{1}{x}}}+x}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{x(\sqrt{1-\frac{1}{x}}-1)}{\sqrt{\sqrt{1-\frac{1}{x}}}+1}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{1-\frac{1}{x}}-1}{\frac{1}{x}[\sqrt{\sqrt{1-\frac{1}{x}}}+1]}$$$$[\frac{1}{x}=t\wedge t\to 0]$$$$=\underset{t\to 0}{\lim }\frac{\sqrt{1-t}-1}{t[\sqrt{\sqrt{1-t}}+1]}$$$$=\frac{1}{2}\times \underset{t\to 0}{\lim }\frac{-t}{t[\sqrt{1-t}+1]}$$$$=-\frac{1}{4}$$

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