Q:

$ \lim_{n \to +\infty} \tfrac{n}{\log n} (\sqrt[n]{2n} -1) $? 

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?  

A:

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

