Let $h(x) = \sqrt{x + \sqrt{x}}  \sqrt{x}$ , find $\lim_{x \to \infty } h(x)$ . I've tried substitution , multiplying by conjugate and l'hospital's rule but didn't work .
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Multiplying and dividing by the "conjugate" $\sqrt{x + \sqrt{x}} + \sqrt{x}$ is a good strategy!
Note that for $x>0$, $$ \sqrt{x + \sqrt{x}}  \sqrt{x}=\frac{ (x + \sqrt{x})  x}{ \sqrt{x + \sqrt{x}} + \sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}} + \sqrt{x}}=\frac{ 1}{ \sqrt{1 + \frac{1}{\sqrt{x}}} + 1}.$$ Can you take it from here?
Robert Z
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Yes , Thanks a lot . – S.H.W Dec 21 '17 at 09:50
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Note that $\sqrt{x+\sqrt{x}}  \sqrt{x} = \frac{x+\sqrt{x}x}{\sqrt{x+\sqrt{x}}+\sqrt{x}} = \frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}} = \frac{1}{\sqrt{1+\sqrt{\frac{1}{x}}}+1}$, and that $\frac{1}{x}$ is tending towards $0$.
This should lead you to the answer.
Higurashi
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