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I suspect that $$\lim_{x \to \infty} \frac{1000^x}{x^x} = 0.$$ However, I do not know how to prove that this is the case. Any help would be greatly appreciated.

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Hint: write ${1000^x\over x^x}= ( {1000\over x})^x$. Note that $0<{1000\over x} $ and that eventually, ${1000\over x}<{1\over 2}$. Then compare (use the squeeze theorem) with ${1\over 2^x}$.

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Write $$\frac{1000^x}{x^x} = \exp(x (\ln 1000 - \ln x))$$ What can you say about $\ln 1000 - \ln x$, and then about $x (\ln 1000 - \ln x)$, as $x \to +\infty$?

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If $x > 2000$ then $\frac{1000^x}{x^x} < \frac{1}{2^x}$.

This echoes an answer I gave recently to another problem. Not sure what that means.

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