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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    \[\lim(x->infinity) \sqrt{x-\sqrt{x-\sqrt{x}}}-\sqrt{x}\]

  2. anonymous
    • 5 years ago
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    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,\[\lim_{n->\infty}\sqrt{x-\sqrt{x-\sqrt{x}}}-\lim_{n->\infty}\sqrt{x}\]\[\iff\lim_{n->\infty}\sqrt{x}-\lim_{n->\infty}\sqrt{x}=0\]

  3. anonymous
    • 5 years ago
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    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.

  4. nikvist
    • 5 years ago
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    answer is \[-\frac{1}{2}\]

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  5. anonymous
    • 5 years ago
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    thanks for the response.. i'm understand now.. :)

  6. anonymous
    • 5 years ago
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    ah, yes

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