$$\large \lim_{x\to\infty} \frac{x^{\log_2(\log_2 x)}}{x^3}$$
How can I evaluate the above limit?
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1$$\lim_{x\to\infty}x^{\log_2(\log_2x)-3}$$ Clearly, $x,\log_2(\log_2x)-3\to\infty$ – lab bhattacharjee Sep 17 '15 at 15:58
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@user271017 Please check my edit. I wasn't sure if you meant to have $\dfrac{\log_2(\log_2x)}{3}$ in the exponent or not. (This will have an effect on your answer.) – user170231 Sep 17 '15 at 15:58
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2Hint:$\forall x>256,\log_2(\log_2(x))>3$ – Aditya Hase Sep 17 '15 at 15:58
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2so it will be ∞ – Arshad Sep 17 '15 at 16:04
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this limit doesn't exist – Dr. Sonnhard Graubner Sep 17 '15 at 16:53
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You may write it as, $$\lim_{x\to\infty}\dfrac{x^{\log_2(\log_2x)}}{x^3}=\lim_{x\to\infty}x^{\log_2(\log_2x)-3}.$$ Obviously, $\log_2(\log_2 x)-3$ will tend to infinity as $x\to\infty$, hence the limit is $\infty$.
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