I am working on an asymptotic analysis question from a data structures past paper, and I need to show that $$\lim\limits_{n\to\infty}\frac{n}{(\log n)^2}=\infty$$
Could I have a hint for working out how to show this, please?
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Hint: Use L'hopital's rule on the limit since it's in an indeterminate form.
user222031
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I have tried to use your hint, but am struggling to make progress. Let $f(n) = n$ and $g(n) = (\log n)^2$ (where we are working base 2). Note that $g(n) = (\frac{\ln(n)}{\ln(2)})^2$. Then $\lim_{n\to\infty}\frac{f(n)}{g(n)} = \lim_{n\to\infty}\frac{f'(n)}{g'(n)} = \lim_{n\to\infty}\frac{1}{2\ln(n)\cdot (\ln2)^{-2}} = 0 \not= \infty.$ Have I done something wrong? – Caleb Owusu-Yianoma May 16 '15 at 18:00
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You should have $g'(n) = \frac{2\ln(n)}{n(\ln(2))^2}$. $$((\ln(n))^2)' =2(\ln(n))(\ln(n))' = \frac{2\ln(n)}{n}$$ – user222031 May 16 '15 at 18:31
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Of course. You correctly identified my mistake. I have solved the problem now. :) – Caleb Owusu-Yianoma May 16 '15 at 18:53
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note that $ log(n) \leq n^\alpha $ for any $\alpha >0$ for instance chose $\alpha = 1/4$ now apply comparison test
Alter: Use L'Hospitals Rule
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How can I show that $\log n \leq n^\alpha \space \forall \alpha \gt 0$? – Caleb Owusu-Yianoma May 16 '15 at 16:55
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Hint: $$\frac{\sqrt n}{\log n}\to +\infty \implies \frac{n}{(\log n)^2} \to +\infty$$ And you can have many ways to prove the first part.
Vim
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