lim-x-x-4-x-3-x-

Question Number 174258 by blackmamba last updated on 28/Jul/22

\lim_{x \to \infty} \sqrt{\sqrt{x^{4} - x^{3}}} - x = ?

Answered by cortano1 last updated on 28/Jul/22

= \lim_{x \to \infty} \frac{\sqrt{x^{4} - x^{3}} - x^{2}}{\sqrt{\sqrt{x^{4} - x^{3}}} + x}

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

= \lim_{x \to \infty} \frac{x (\sqrt{1 - \frac{1}{x}} - 1)}{\sqrt{\sqrt{1 - \frac{1}{x}}} + 1}

= \lim_{x \to \infty} \frac{\sqrt{1 - \frac{1}{x}} - 1}{\frac{1}{x} [\sqrt{\sqrt{1 - \frac{1}{x}}} + 1]}

[\frac{1}{x} = t \land t \to 0]

= \lim_{t \to 0} \frac{\sqrt{1 - t} - 1}{t [\sqrt{\sqrt{1 - t}} + 1]}

= \frac{1}{2} \times \lim_{t \to 0} \frac{- t}{t [\sqrt{1 - t} + 1]}

= - \frac{1}{4}

Facebook Tweet Pin