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

Question

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?

As per this question, we know that $\lim_{n\to\infty}n(\sqrt[n]x-1)=\ln(x)$, thus, what can you say about your limit? — Simply Beautiful Art, Feb 23 '17 at 01:44
I can't understand how to use your suggestion, could you show me that? — ambimath, Feb 23 '17 at 01:53
Your limit is of the following form: $$\frac c\infty$$Since we have a finite part divided by an unbounded part. — Simply Beautiful Art, Feb 23 '17 at 01:54
@SimplyBeautifulArt : the limit is not of the form $c/\infty$. The right answer is $1$ as given by zhw. — Paramanand Singh, Feb 23 '17 at 03:34

score 1 · Accepted Answer · answered Feb 23 '17 at 03:28

1

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

answered Feb 23 '17 at 03:28

zhw.

1 Answers1