real analysis - Limit $limlimits_{ntoinfty} sqrt[n]{frac1{sqrt3}left(left(frac{1+sqrt3}2right)^n-left(frac{1-sqrt3}2right)^nright)}$

I have problems with finding: $$\lim_{n\to\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n}-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$$ I tried to do it following way:

$\displaystyle\lim_{n\to\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n}-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$

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

$1\cdot\displaystyle\lim_{n\to\infty} \sqrt[n]{\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$

Now I used formula for difference of powers:

$\Big(\frac{1+\sqrt{3}}{2}\Big)^n-\Big(\frac{1-\sqrt{3}}{2}\Big)^n=\Big(\frac{1+\sqrt{3}}{2}-\frac{1-\sqrt{3}}{2}\Big)\cdot\Big( \Big(\frac{1+\sqrt{3}}{2}\Big)^{n-2}\Big(\frac{1+\sqrt{3}}{2}\Big)+\Big(\frac{1+\sqrt{3}}{2}\Big)^{n-3}\Big(\frac{1+\sqrt{3}}{2}\Big)\Big(\frac{1-\sqrt{3}}{2}\Big)+\Big(\frac{1+\sqrt{3}}{2}\Big)^{n-4}\Big(\frac{1+\sqrt{3}}{2}\Big)\Big(\frac{1-\sqrt{3}}{2}\Big)+...+\Big(\frac{1-\sqrt{3}}{2}\Big)^{n-1}\Big)$

In the last parenthesis, I saw two geometric series and I tried to add them, however, it quickly appeared that there will be other geometric series and here is where I am a bit helpless (it is getting very nasty very quickly). Do you have any hints to move it in maybe another way? I would be very grateful, thanks!

Answer

For the beginning prove the following facts
$$\lim\limits_{n\to\infty} \sqrt[n]{a}=1 \quad\text{ for }\quad a>0$$
$$\lim\limits_{n\to\infty} \sqrt[n]{x^n-y^n}=x\quad\text{ for }\quad x>|y|$$
then apply them to the limit
$$\lim\limits_{n\to\infty}\sqrt[n]{a(x^n-y^n)}$$