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This limit wildly appeared in a test and, well, I could not solve it without using L'Hôspital rule - which were not allowed in the test. Can anyone help me?

$$\lim_{x \to +\infty}\sqrt{x-\sqrt{x}}-\sqrt{x-1}$$


Answer, thanks to @user71352. $$ \lim_{x\to\infty}\frac{\sqrt{x-\sqrt{x}}-\sqrt{x-1}}{1}\cdot \frac{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}} $$ Multiply by the conjugate: $$ \lim_{x\to\infty}\frac{(\sqrt{x-\sqrt{x}})^2-(\sqrt{x-1})^2}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}} $$ Multiply by $\frac{1}{\sqrt{x}}$: $$ \lim_{x\to\infty}\frac{1-\sqrt{x}}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}\cdot\frac{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x}}} $$ Knowing that $\lim_{x\to\infty}\frac{1}{x} = \lim_{x\to\infty}\frac{1}{\sqrt{x}} = 0$: $$ \lim_{x\to\infty}\frac{\frac{1}{\sqrt{x}}-1}{\sqrt{1-\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{x}}}=\frac{0-1}{\sqrt{1-0}+\sqrt{1-0}}=-\frac{1}{2} $$

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    $\begingroup$ Multiply by the "conjugate." $\endgroup$ – symplectomorphic May 1 '14 at 2:47
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$$ \sqrt{x-\sqrt{x}}-\sqrt{x-1}=\frac{x-\sqrt{x}-x+1}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}=\frac{1-\sqrt{x}}{\sqrt{x-\sqrt{x}}+\sqrt{x-1}}=\frac{\frac{1}{\sqrt{x}}-1}{\sqrt{1-\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{x}}} $$

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  • $\begingroup$ You're welcome. $\endgroup$ – user71352 May 1 '14 at 3:46

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