lim(x->infinity) sqrt(x-sqrt(x-sqrt(x)))-sqrt(x)

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lim(x->infinity) sqrt(x-sqrt(x-sqrt(x)))-sqrt(x)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\lim(x->infinity) \sqrt{x-\sqrt{x-\sqrt{x}}}-\sqrt{x}\]
Hello, I'm thinking you might want to try asymptotic equivalence to show first that\[x-\sqrt{x} \]~\[x\] (i.e. x-sqrt{x} is asymptotically equivalent to x). The two are as.eq. since\[\lim_{x-> \infty}\frac{x-\sqrt{x}}{x}=\lim_{x-> \infty}\frac{1-x^{-1/2}}{1}=1\]That means, as x approaches infinity, \[\sqrt{x-\sqrt{x-\sqrt{x}}} \iff \sqrt{x-\sqrt{x}}\iff \sqrt{x}\] which means,\[\lim_{n->\infty}\sqrt{x-\sqrt{x-\sqrt{x}}}-\lim_{n->\infty}\sqrt{x}\]\[\iff\lim_{n->\infty}\sqrt{x}-\lim_{n->\infty}\sqrt{x}=0\]
Where it should be noted that the logical equivalence symbol \[\iff\]should really be ~ since we're dealing with equivalence relations. I couldn't use the proper symbol in the equation editor for some reason.

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answer is \[-\frac{1}{2}\]
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thanks for the response.. i'm understand now.. :)
ah, yes

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