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Question Number 117052 by bemath last updated on 09/Oct/20

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

Answered by bobhans last updated on 09/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{4\mathrm{x}(1+\frac{\sqrt{4\mathrm{x}+\sqrt{4\mathrm{x}+\sqrt{4\mathrm{x}}}}}{4\mathrm{x}})}-\sqrt{4\mathrm{x}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{4\mathrm{x}}[\sqrt{1+\sqrt{\frac{1}{4\mathrm{x}}+\sqrt{\frac{1}{{\left(4\mathrm{x}\right)}^3}+\sqrt{\frac{1}{{\left(4\mathrm{x}\right)}^7}}}}}-1]=$$$$\mathrm{letting}\sqrt{4\mathrm{x}}=\frac{1}{\mathrm{t}}\to \{\begin{array}{l}\frac{1}{4\mathrm{x}}={\mathrm{t}}^2\\ \frac{1}{{\left(4\mathrm{x}\right)}^3}={\mathrm{t}}^6\\ \frac{1}{{\left(4\mathrm{x}\right)}^7}={\mathrm{t}}^{14}\end{array}$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{\sqrt{1+\sqrt{{\mathrm{t}}^2+\sqrt{{\mathrm{t}}^6+\sqrt{{\mathrm{t}}^{14}}}}}-1}{\mathrm{t}}=$$$$\frac{1}{2}\times \underset{\mathrm{t}\to 0}{\lim }\frac{\sqrt{{\mathrm{t}}^2+\sqrt{{\mathrm{t}}^6+{\mathrm{t}}^7}}}{\mathrm{t}}=\frac{1}{2}\times \underset{\mathrm{t}\to 0}{\lim }\frac{\sqrt{{\mathrm{t}}^2+{\mathrm{t}}^3\sqrt{1+\mathrm{t}}}}{\mathrm{t}}$$$$=\frac{1}{2}\times \underset{\mathrm{t}\to 0}{\lim }\frac{\mathrm{t}\sqrt{1+\mathrm{t}\sqrt{1+\mathrm{t}}}}{\mathrm{t}}=\frac{1}{2}$$

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