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I've tried so hard to find this limit, could you help me? $ \lim_{n \to +\infty} \tfrac{n}{\log n} (\sqrt[n]{2n} -1) $ . Do you have any suggestion?

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    As per [this question](http://math.stackexchange.com/questions/2157147/elegant-solution-to-lim-limits-n-to-inftyn-sqrtna-1), we know that $\lim_{n\to\infty}n(\sqrt[n]x-1)=\ln(x)$, thus, what can you say about your limit? – 2017-02-23
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    I can't understand how to use your suggestion, could you show me that? – 2017-02-23
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    Your limit is of the following form: $$\frac c\infty$$Since we have a finite part divided by an unbounded part. – 2017-02-23
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    uuuuu ok, now i undestand! Thanks you :) – 2017-02-23
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    No problem! :-) – 2017-02-23
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    @SimplyBeautifulArt I'm not sure I understand your comment. – 2017-02-23
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    @SimplyBeautifulArt : the limit is not of the form $c/\infty$. The right answer is $1$ as given by zhw. – 2017-02-23

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Hint: The expression equals

$$\frac{n}{\ln n}(e^{(\ln 2n)/n}-1) = \frac{n}{\ln n}(\ln 2n)/n)\frac{e^{(\ln 2n)/n}-1}{(\ln 2n)/n}.$$

Recall $\lim_{u\to 0} (e^u-1)/u = 1.$