$\lim_{n\to \infty} n(\alpha -\log(\log n))$

Question

Can someone show why this limit is $0$? $$\lim_{n\to \infty} \frac{n^\alpha}{(\log n)^{\log n}}=\lim_{n\to \infty} \frac{e^{\alpha \log n}}{e^{\log n \cdot\log(\log n)}}=\lim_{n\to \infty} e^{\alpha \log n-\log n\cdot \log(\log n)}=\lim_{n\to \infty} e^{\log n(\alpha -\log(\log n))}=\lim_{n\to \infty} n(\alpha -\log(\log n))$$ with $\alpha>0$

I tried with L'Hôpital's rule but at a certain moment the limit becomes $+ \infty$, so I think it is that and not $0$

Answer 1

Note that

$$\frac{n^\alpha}{(\log n)^{\log n}}=e^{\log{\frac{n^\alpha}{(\log n)^{\log n}}}} \to e^{-\infty}=0$$

indeed

$$\log{\frac{n^\alpha}{(\log n)^{\log n}}}=\alpha\log n-\log n\cdot\log\log n=\log n(\alpha-\log\log n)\to-\infty$$

Answer 2

Your last step is wrong. You wrote

$$e^{\log n \cdot (\text{stuff})} = n \cdot (\text{stuff})$$

which is not a valid usage of the properties of exponents and logs. The correct conclusion is

$$e^{\log n \cdot (\text{stuff})} = n^{(\text{stuff})}$$

Keeping this in mind, you have

$$\lim_{n \to \infty} n^{\log n (\alpha - \log \log n)}$$

Now the exponent clearly tends to $-\infty$, and since $n \to \infty$ the overall limit is zero.

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Alternatively, you could have noted at the start that

$$(\log n)^{\log n} = e^{\log n \log \log n} = n^{\log \log n}.$$

The exponent here grows without bound, so the denominator is vastly larger than the numerator - thus, the limit is zero.

Answer 3

ok now I see that the limit becomes

$$\lim_{n\to \infty} n^{\alpha -\log(\log n)}$$ with $\alpha>0$

but the form is indeterminated and the hopital rule complicates the things