How to solve this limit, and what did I do wrong?

Question

I have to solve: $$ \lim_{n\to\infty}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1} $$

Here's my attempt:

$$ \lim_{n\to\infty}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1} = \lim_{n\to\infty}\frac{(1+\frac{2}{n+1})^{1/6}-1}{\frac{2}{n+1}}\cdot\frac{2}{(n+1)(\sqrt[n]{2}-1)}=\frac{1}{6}\lim_{n\to\infty}\frac{2}{\frac{2^{1/n}-1}{1/n}+\sqrt[n]{2}-1}=\frac{2}{6(\ln2-1)}=\frac{1}{3\ln2-3}$$

But the result is supposed to be $\frac{1}{3\ln2}$, what did I do wrong?

PS: I used the following limits in my attempt to solve it:

$$\lim_{n\to\infty}\frac{(1+x_n)^r-1}{x_n}=r$$ $$\lim_{n\to\infty}\frac{a^{x_n}-1}{x_n}=\ln a$$

Answer 1

From here

$$\ldots=\frac{1}{6}\lim_{n\to\infty}\frac{2}{\frac{2^{1/n}-1}{1/n}+\sqrt[n]{2}-1}=\frac{1}{6}\frac{2}{\ln 2+1-1}=\ldots$$

As an alternative

$$\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1}=\frac{\frac1n}{\sqrt[n]{2}-1}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\frac1n}\to \frac{1}{3\ln2}$$

indeed

• $\frac{\frac1n}{\sqrt[n]{2}-1}\to \frac{1}{\ln2}$

and by binomial approximation

• $\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\frac1n}=n\left(\left(1+\frac{2}{n+1}\right)^\frac16-1\right) \approx n\left(1+\frac{2}{6n+6}-1\right)=\frac{n}{3n+3}\to \frac13$

Answer 2

You made your mistake when you claimed $$\lim\limits_{n\to\infty}\dfrac{(1+\frac{2}{n+1})^{1/6}-1}{\frac{2}{n+1}}\cdot \dfrac{2}{(n+1)(\sqrt[n]{2}-1)}=\dfrac{1}{6}\lim\limits_{n\to\infty}\dfrac{2}{\frac{2^{1/n}-1}{1/n}+\sqrt[n]{2}-1}$$$$\\=\dfrac{2}{6(\ln2-1)}$$

because you assumed that $\lim\limits_{n\to\infty} (\sqrt[n]{2}-1)=-1$ when it actually equals $0.$ Fix that and you'll get the right answer.