Exploring $\lim_{n\rightarrow \infty }(2-2^{\frac{1}{n+1}})^{n}$

[Vali]

$$\lim_{n\rightarrow \infty }(2-2^{\frac{1}{n+1}})^{n}$$
I wrote that $L=e^{\lim_{n\rightarrow \infty }(n)(1-2^{\frac{1}{n+1}})}$
how to continue ?

 

[MarkFL]

What I would do is write:

\(\displaystyle L=\lim_{n\to\infty}\left(\left(2-2^{\frac{1}{n+1}}\right)^n\right)\)

And this implies:

\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\frac{\ln\left(2-2^{\frac{1}{n+1}}\right)}{\dfrac{1}{n}}\right)\)

Now you have the indeterminate form 0/0.

 

[Vali]

After I apply L'Hopital I get $ln(L)=-ln(2)$ so $L=\frac{1}{2}$
One question,Can this limit be solved by using this remarkable limit? $\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e$
Thank you very much!