lim-x-x-x-2-2x-x-2-x-2-x-

Question Number 159229 by tounghoungko last updated on 14/Nov/21

\lim_{x \to \infty} x (\sqrt{x^{2} + 2 x} + x - 2 \sqrt{x^{2} + x}) = ?

Answered by FongXD last updated on 14/Nov/21

L = \underset{x \to + \infty}{\lim x} (\sqrt{x^{2} + 2 x} - x + 2 x - 2 \sqrt{x^{2} + x})

L = \underset{x \to + \infty}{\lim x} [\frac{(x^{2} + 2 x) - x^{2}}{\sqrt{x^{2} + 2 x} + x} + \frac{{4 x}^{2} - 4 (x^{2} + x)}{2 x + 2 \sqrt{x^{2} + x}}]

L = \underset{x \to + \infty}{\lim x} (\frac{2 x}{\sqrt{x^{2} + 2 x} + x} - \frac{2 x}{x + \sqrt{x^{2} + x}})

L = {\underset{x \to + \infty}{\lim 2 x}}^{2} [\frac{(x + \sqrt{x^{2} + x}) - (\sqrt{x^{2} + 2 x} + x)}{(\sqrt{x^{2} + 2 x} + x) (x + \sqrt{x^{2} + x})}]

L = {\underset{x \to + \infty}{\lim 2 x}}^{2} [\frac{(x^{2} + x) - (x^{2} + 2 x)}{(\sqrt{x^{2} + 2 x} + x) (x + \sqrt{x^{2} + x}) (\sqrt{x^{2} + x} + \sqrt{x^{2} + 2 x})}]

L = \lim_{x \to + \infty} \frac{- {2 x}^{3}}{x^{3} (\sqrt{1 + {2 x}^{- 1}} + 1) (1 + \sqrt{1 + x^{- 1}}) (\sqrt{1 + x^{- 1}} + \sqrt{1 + {2 x}^{- 1}})}

L = \frac{- 2}{(\sqrt{1 + 0} + 1) (1 + \sqrt{1 + 0}) (\sqrt{1 + 0} + \sqrt{1 + 0})} = - \frac{1}{4}

Answered by qaz last updated on 15/Nov/21

\underset{x \to \infty}{\lim x} (\sqrt{x^{2} + 2 x} + x - 2 \sqrt{x^{2} + x})

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

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

= \lim_{x \to 0, ξ \to 4} \frac{1}{x^{2}} \cdot \frac{1}{2 \sqrt{ξ}} ({(\sqrt{1 + 2 x} + 1)}^{2} - (4 + 4 x))

= \frac{1}{2} \lim_{x \to 0} \frac{\sqrt{1 + 2 x} - x - 1}{x^{2}}

= \frac{1}{2} \lim_{x \to 0, ζ \to 1} \frac{1}{x^{2}} \cdot \frac{1}{2 \sqrt{ζ}} (1 + 2 x - {(x + 1)}^{2})

= - \frac{1}{4}

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