Evaluate-lim-x-0-x-1-x-x-1-x-2-x-x-1-1-

Question Number 173729 by a.lgnaoui last updated on 17/Jul/22

E v a l u a t e

{l i m}_{x \to 0} \frac{x - \frac{1}{\sqrt{x} - x} + 1}{x^{2} + \frac{\sqrt{x}}{\sqrt{x} - 1} - 1}

Answered by blackmamba last updated on 17/Jul/22

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

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

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

= \infty

Answered by greougoury555 last updated on 17/Jul/22

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

[l e t \frac{1}{\sqrt{x} - 1} = t; \sqrt{x} = \frac{1 + t}{t}]

= \lim_{t \to - 1} (\frac{{(\frac{1 + t}{t})}^{2} + 1 + \frac{t^{2}}{1 + t}}{[{(\frac{1 + t}{t})}^{2} - 1] [{(\frac{1 + t}{t})}^{2} + 1] + 1 + t})

= - \infty

Commented by a.lgnaoui last updated on 17/Jul/22

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