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As above, I have to calculate $\lim_{n\rightarrow\infty} n\left ( \ln(n^2 +1) -2\ln(n) \sqrt[n]{\ln(n)} \right )$I tried to multiply it by $ \frac{\left ( \ln(n^2 +1) +2\ln(n)\sqrt[n]{\ln(n)}\right )}{\left ( \ln(n^2 +1) +2\ln(n)\sqrt[n]{\ln(n)}\right )}$ so that it is $\lim_{n\rightarrow\infty} n \frac{\left ( \ln^2(n^2 +1) -2\ln^2 (n) \sqrt[\frac{1}{2}n]{\ln(n)} \right )}{\left ( \ln(n^2 +1) +2\ln(n)\sqrt[n]{\ln(n)}\right )},$but it is not so easy to play with (first glance says that it goes to infinity), nevertheless, can somebody take a closer look at this and suggest something, please? I would be very grateful.

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The following expansions yield readily an equivalent, hence the limit:

  • $\ln(n^2+1)=2\log(n)+\log\left(1+\frac1{n^2}\right)=2\log(n)+o\left(\frac1n\right)$.
  • $\sqrt[n]{\log n}=\exp\left(\frac1n\log\log n\right)=1+\frac1n\log(\log n)+o\left(\frac1n\right)$.