lim-r-4r-2-2r-8r-3-4r-2-1-3-

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

\lim_{r \to \infty} \sqrt{4 r^{2} + 2 r} - \sqrt[3]{8 r^{3} + 4 r^{2}} = ?

Answered by bemath last updated on 14/Sep/20

\Leftrightarrow \lim_{r \to \infty} r (\sqrt{4 + \frac{2}{r}} - \sqrt[3]{8 + \frac{4}{r}}) =

[s e t \frac{1}{r} = b, b \to 0]

\lim_{b \to 0} \frac{\sqrt{4 + 2 b} - \sqrt[3]{8 + 4 b}}{b} =

\lim_{b \to 0} \frac{2 \sqrt{1 + \frac{b}{2}} - 2 \sqrt[3]{1 + \frac{b}{2}}}{b} =

2 \times \lim_{b \to 0} \frac{(1 + \frac{b}{4}) - (1 + \frac{b}{6})}{b} =

2 \times \lim_{b \to 0} \frac{(\frac{3 b - 2 b}{12})}{b} = 2 \times \frac{1}{12} = \frac{1}{6}

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