lim-x-8-x-3-x-1-3-4-x-2-x-

Question Number 153404 by liberty last updated on 07/Sep/21

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

Answered by EDWIN88 last updated on 07/Sep/21

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

L_{1} = \lim_{x \to \infty} \frac{(8^{x} + 3^{x}) - 8^{x}}{\sqrt[3]{{(8^{x} + 3^{x})}^{2}} + \sqrt[3]{8^{x} (8^{x} + 3^{x})} + 4^{x}}

L_{1} = \lim_{x \to \infty} \frac{3^{x}}{\sqrt[3]{8^{2 x} {(1 + {(\frac{3}{8})}^{x})}^{2}} + \sqrt[3]{8^{2 x} (1 + {(\frac{3}{8})}^{x})} + 4^{x}}

L_{1} = \lim_{x \to \infty} \frac{{(\frac{3}{4})}^{x}}{\sqrt[3]{{(1 + {(\frac{3}{8})}^{x})}^{2}} + \sqrt[3]{1 + {(\frac{3}{8})}^{x}} + 1}

L_{1} = \frac{0}{3} = 0

L_{2} = \lim_{x \to \infty} \sqrt{4^{x}} - \sqrt{4^{x} - 2^{x}}

L_{2} = \lim_{x \to 0} 2^{x} (1 - \sqrt{1 - {(\frac{1}{2})}^{x}})

L_{2} = \lim_{X \to 0} \frac{1 - \sqrt{1 - X}}{X} = \frac{1 - (1 - \frac{1}{2} X)}{X} = \frac{1}{2}

T h e n L = L_{1} + L_{2} = 0 + \frac{1}{2} = \frac{1}{2}

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