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

$$Evaluate$$$${lim}_{x\to 0}\frac{x-\frac{1}{\sqrt{x}-x}+1}{x^2+\frac{\sqrt{x}}{\sqrt{x}-1}-1}$$

Answered by blackmamba last updated on 17/Jul/22

$$L=\underset{x\to 0}{\lim }\left(\frac{x-\frac{1}{\sqrt{x}-x}-1}{x^2+\frac{\sqrt{x}}{\sqrt{x}-1}-1}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{(x-1)+\frac{1}{\sqrt{x}(\sqrt{x}-1)}}{(x-1)(x+1)+\frac{\sqrt{x}}{\sqrt{x}-1}}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}-1}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{{(\sqrt{x}-1)}^2(\sqrt{x}+1)+\frac{1}{\sqrt{x}}}{{(\sqrt{x}-1)}^2(\sqrt{x}+1)(x+1)+\sqrt{x}}\right)$$$$=\mathrm{\infty }$$$$$$

Answered by greougoury555 last updated on 17/Jul/22

$$=\underset{x\to 0}{\lim }\left(\frac{(x+1)+\frac{1}{\sqrt{x}(\sqrt{x}-1)}}{(x-1)(x+1)+\frac{\sqrt{x}}{\sqrt{x}-1}}\right)$$$$[let\frac{1}{\sqrt{x}-1}=t;\sqrt{x}=\frac{1+t}{t}]$$$$=\underset{t\to -1}{\lim }\left(\frac{{\left(\frac{1+t}{t}\right)}^2+1+\frac{t^2}{1+t}}{[{\left(\frac{1+t}{t}\right)}^2-1][{\left(\frac{1+t}{t}\right)}^2+1]+1+t}\right)$$$$=-\mathrm{\infty }$$

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

$$thanksforyou$$$$$$$$$$

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