How to find $\lim_{n\to\infty}(2-\sqrt[n]{2})^n $? $$\lim_{n\to\infty}(2-\sqrt[n]{2})^n $$ 
I have tried the following:
Let $ a_n= (2-\sqrt[n]{2})^n$. Then
$$\ln(a_n)= n \ln(2-\sqrt[n]{2})=n \cdot(1-2^\frac{1}{n})\cdot \frac{ \ln (1+(1-2^\frac{1}{n})}{(1-2^\frac{1}{n})} $$ 
The latter goes to  $1$ and I don't know how to do with the rest. I don't even know if what I am doing is actually mathematically correct. 
 A: Using the Taylor expansion $2^x=1+x\log2+o(x)$, we have
$$(2-2^{1/n})^n=\left(1-\frac{\log2}{n}+o(\tfrac1n)\right)^n\to e^{-\log2}=\frac12.$$
A: $$\lim_{n\to\infty}(2-2^{\frac{1}{n}})^n = \lim_{n\to\infty}(1+1-2^{\frac{1}{n}})^n = \lim_{n\to\infty}\bigg[\big(1+1-2^{\frac{1}{n}}\big)^{\frac{1}{1-2^{\frac{1}{n}}}}\bigg]^{n\cdot(1-2^{\frac{1}{n}})} = e^{\lim_{n\to\infty}\frac{1-2^{\frac{1}{n}}}{\frac{1}{n}}} = e^{-\ln2} = \frac{1}{2}$$
Using $$\lim_{x\to 0}(1+x)^{\frac{1}{x}} = e$$ and $$\lim_{x\to0}\frac{a^x-1}{x} = \ln a$$
A: You can rewrite $(2-2^{1/n})^n$ as $\exp(n\log(2-2^{1/n})$ and because $\exp$ is continuous we know that $$\lim_{n\to\infty}\exp(f(x))=\exp\left(\lim_{n\to\infty}f(x)\right)$$
We also know that $$\lim_{x\to0} \frac{\log(x+1)}{x}=1$$
Since $2-2^{1/n}\to1$ as $n\to\infty$ we can multiply numerator and denominator of our expression by $1-2^{1/n}$ getting $$\lim_{n\to\infty}n\log\left(2-2^{1/n}\right)=\lim_{n\to\infty}-n\left(2^{1/n}-1\right)$$
which is equal to $-\log(2)$ because $$\lim_{n\to\infty}n\left(a^{1/n}-1\right)=\log(a)$$
So your limit is equal to $\exp\log(2^{-1})=\frac{1}{2}$
