Limit of two logarithms subtraction

Question

I have the problem with calculating such limit:

$$\lim_{n \to \infty} \frac{n^{\ln(n)}}{2^n}$$

After some transformations, I managed to get it to:

$$e^{\lim_{n\to\infty}((\ln n)^2 - n\ln(2))}$$

though I can't do anything more than that. I kind of know the result from the beginning, but I can't get past this moment. As there is a difference in the limit, I'm not sure how should I deal with it. I think I know that $n\ln(2)$ grows asymptotically faster, but it isn't enough proof to say the limit of that subtraction is $-\infty$, and, for the first limit: $e^{-\infty} = 0$.

Long story short: how should I solve this part? $$\lim_{n\to\infty}((\ln n)^2 - n\ln(2))$$

Answer 1

HINT

$$(\ln n)^2 - n\ln 2=(\ln n)^2\left(1-\frac{n\ln 2}{(\ln n)^2}\right)$$

Answer 2

For any $k$, $$ \begin{align} \frac{\log(n)}n &=\frac1n\int_1^n\frac1x\,\mathrm{d}x\\ &=\frac1n\int_1^k\frac1x\,\mathrm{d}x+\frac1n\int_k^n\frac1x\,\mathrm{d}x\\[3pt] &\le\frac{\log(k)}n+\frac1k\frac{n-k}n\tag1 \end{align} $$ Taking the limit of $(1)$, we get $$ \lim_{n\to\infty}\frac{\log(n)}n\le\frac1k\tag2 $$ Since $(2)$ is true for any $k$, we must have $$ \lim_{n\to\infty}\frac{\log(n)}n=0\tag3 $$ Therefore, $$ \begin{align} \lim_{n\to\infty}\frac{\log(n)^2}n &=\lim_{n\to\infty}\left(\frac{2\log\left(\sqrt{n}\right)}{\sqrt{n}}\right)^2\\ &=4\,\left(\lim_{n\to\infty}\frac{\log\left(\sqrt{n}\right)}{\sqrt{n}}\right)^2\\[6pt] &=4\cdot0^2\\[15pt] &=0\tag4 \end{align} $$

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You can now use $(4)$ to evaluate $$ \lim_{n\to\infty}\left(\log(n)^2-n\log(2)\right)=\lim_{n\to\infty}n\left(\frac{\log(n)^2}n-\log(2)\right)\tag5 $$