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Question Number 113576 by bemath last updated on 14/Sep/20

$$\underset{r\to \mathrm{\infty }}{\lim }\sqrt{4r^2+2r}-\sqrt[3]{8r^3+4r^2}=?$$

Answered by bemath last updated on 14/Sep/20

$$\iff \underset{r\to \mathrm{\infty }}{\lim }r(\sqrt{4+\frac{2}{r}}-\sqrt[3]{8+\frac{4}{r}})=$$$$[set\frac{1}{r}=b,b\to 0]$$$$\underset{b\to 0}{\lim }\frac{\sqrt{4+2b}-\sqrt[3]{8+4b}}{b}=$$$$\underset{b\to 0}{\lim }\frac{2\sqrt{1+\frac{b}{2}}-2\sqrt[3]{1+\frac{b}{2}}}{b}=$$$$2\times \underset{b\to 0}{\lim }\frac{(1+\frac{b}{4})-(1+\frac{b}{6})}{b}=$$$$2\times \underset{b\to 0}{\lim }\frac{\left(\frac{3b-2b}{12}\right)}{b}=2\times \frac{1}{12}=\frac{1}{6}$$

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