\displaystyle\lim_{n\to\infty}\frac{\left(\frac{2}{3}\right)^n+\left(\frac{1}{3}\right)^n}{1-\left(\frac{1}{3}\right)^n}=\frac{\displaystyle\lim_{n\to\infty}\Big(\frac{2}{3}\Big)^n+\displaystyle\lim_{n\to\infty}\Big(\frac{1}{3}\Big)^n}{1-\displaystyle\lim_{n\to\infty}\Big(\frac{1}{3}\Big)^n}=\frac{0+0}{1-0}=0