lim-x-2-x-2-4-1-3-x-3-4-x-2-4-x-2-1-3-

Question Number 207816 by efronzo1 last updated on 27/May/24

\lim_{x \to 2} \frac{\sqrt[3]{x^{2} + 4} - \sqrt{x^{3} - 4}}{\sqrt{x^{2} - 4} - \sqrt[3]{x - 2}}

Commented by Frix last updated on 27/May/24

I think {it}^{'} s 0

Answered by Berbere last updated on 28/May/24

x = 2 + y

= \lim_{y \to 0} \frac{\sqrt[3]{8 + 4 y + 4 y^{2}} - \sqrt{4 + 12 y + 6 y^{2} + y^{3}}}{\sqrt{4 y + y^{2}} - \sqrt[3]{y}}

= \lim_{y \to 0} \frac{\frac{1}{2} (\sqrt{1 + \frac{y}{2} + \frac{y^{2}}{2}} - \sqrt{1 + \frac{y^{3}}{4} + \frac{3 y^{2}}{2} + 3 y})}{\sqrt{y} + \sqrt{4 y + y^{2}}}

\sqrt{1 + \frac{y}{2} + \frac{y^{2}}{2}} - \sqrt{1 + 3 y + \frac{y^{3}}{4} + \frac{3 y^{2}}{2}} = 1 + \frac{1}{2} (\frac{y}{2}) - 1 - \frac{1}{3} (\frac{y}{2}) + o (y) = \frac{y}{12} + o (y)

= \lim_{y \to 0} \frac{\frac{y}{12}}{- \sqrt[3]{y} + \sqrt{4 y + y^{2}}} = \frac{y^{\frac{2}{3}}}{12 (- 1 + \sqrt{4 y^{\frac{1}{3}} + y^{\frac{4}{3}}}}) = 0

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