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anonymous
 5 years ago
lim(x>infinity) sqrt(xsqrt(xsqrt(x)))sqrt(x)
anonymous
 5 years ago
lim(x>infinity) sqrt(xsqrt(xsqrt(x)))sqrt(x)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim(x>infinity) \sqrt{x\sqrt{x\sqrt{x}}}\sqrt{x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hello, I'm thinking you might want to try asymptotic equivalence to show first that\[x\sqrt{x} \]~\[x\] (i.e. xsqrt{x} is asymptotically equivalent to x). The two are as.eq. since\[\lim_{x> \infty}\frac{x\sqrt{x}}{x}=\lim_{x> \infty}\frac{1x^{1/2}}{1}=1\]That means, as x approaches infinity, \[\sqrt{x\sqrt{x\sqrt{x}}} \iff \sqrt{x\sqrt{x}}\iff \sqrt{x}\] which means,\[\lim_{n>\infty}\sqrt{x\sqrt{x\sqrt{x}}}\lim_{n>\infty}\sqrt{x}\]\[\iff\lim_{n>\infty}\sqrt{x}\lim_{n>\infty}\sqrt{x}=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where 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.

nikvist
 5 years ago
Best ResponseYou've already chosen the best response.0answer is \[\frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for the response.. i'm understand now.. :)
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