Calculate-lim-x-x-3-3x-2-1-3-x-2-2x-lim-x-7-x-2-x-20-1-3-x-9-1-4-2-

Question Number 162672 by LEKOUMA last updated on 31/Dec/21

C a l c u l a t e

\lim_{x \to + \infty} (\sqrt[3]{x^{3} + 3 x^{2}} - \sqrt{x^{2} - 2 x})

\lim_{x \to 7} \frac{\sqrt{x + 2} - \sqrt[3]{x + 20}}{\sqrt[4]{x + 9} - 2}

Answered by tounghoungko last updated on 31/Dec/21

(2) \lim_{x \to 7} \frac{\sqrt{x + 2} - 3 + 3 - \sqrt[3]{x + 20}}{\sqrt[4]{x + 9} - 2}

L_{1} = \lim_{x \to 7} \frac{x - 7}{(\sqrt{x + 2} + 3) (\sqrt[4]{x + 9} - 2)}

L_{1} = \lim_{x \to 7} \frac{(x - 7) 4}{6 (\sqrt{x + 9} - 4)} = \frac{2}{3} \lim_{x \to 7} \frac{8 (x - 7)}{(x - 7)} = \frac{16}{3}

L_{2} = \lim_{x \to 7} \frac{3 - \sqrt[3]{x + 20}}{\sqrt[4]{x + 9} - 2} = \lim_{x \to 7} \frac{(7 - x)}{27} . \lim_{x \to 7} \frac{4}{\sqrt{x + 9} - 4}

L_{2} = \frac{4}{27} . \lim_{x \to 7} \frac{7 - x}{x - 7} . (8) = - \frac{32}{27}

∴ L = L_{1} + L_{2} = \frac{16}{3} - \frac{32}{27} = \frac{112}{27}

Commented by LEKOUMA last updated on 31/Dec/21

T h a n k

Answered by tounghoungko last updated on 31/Dec/21

(1) \lim_{x \to \infty} x \sqrt[3]{1 + \frac{3}{x}} - x \sqrt{1 - \frac{2}{x}}

[\frac{1}{x} = y]

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

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

Commented by LEKOUMA last updated on 31/Dec/21

T h a n k

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