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What is the limit of

$$\lim_{n \to \infty} \frac{\log n}{n^\delta}$$

where $\delta > 0$?

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$0{}{}{}{}{}{}$. – Sabyasachi Mar 19 '14 at 6:54

Hint 1: Substituting $n\mapsto n^{1/\delta}$ yields $$ \lim_{n\to\infty}\frac{\log(n)}{n^\delta}=\lim_{n\to\infty}\frac1\delta\frac{\log(n)}{n} $$ Hint 2: Using the inequality for all real $x$ $$ 1+x\le e^x $$ Substitute $x\mapsto x/2$ and square to get $$ \left(1+\frac x2\right)^2\le e^x $$ which says $$ \frac x{e^x}\le\frac4{2+x} $$ From this, we can deduce that $$ \frac{\log(n)}{n}\le\frac4{2+\log(n)} $$

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$$\lim_{n \to \infty} \frac{\ln n}{n^\delta} = \lim_{n \to \infty} \frac{1}{n \cdot \delta n^{\delta - 1}} = \lim_{n \to \infty} \frac{1}{\delta n^\delta} = 0$$

where in the first passage we used Hopital rule

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Write: $n^a = e^{a\ln n} > 1 + a\ln n + (a \ln n)^{\frac{2}{2!}}$. Then: $\frac{\ln n}{n^a} < \frac{1}{a\ln n} \to 0$.
So the answer is: $0$ (zero)

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$\ln(n)$ grows slower than linear so whatever is $n$ the denumerator grows faster than the numerator and the limit is 0

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Without some calculus it isn't obvious, I suppose, that logarithm grows slower than, for example, $n^{10^{-100000}}$. – sas Mar 19 '14 at 7:03

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