# Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$

Evaluate $$\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$$.
My approach is to do $$\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}} = \lim_{n\to\infty} \frac{\left(\frac{9}{4}\right)^{n}}{\frac{3}{2}\left(\frac{2}{3}\right)^{n}+\frac{3}{2}\left(\frac{9}{4}\right)^{n}}$$ but not sure what to do next.
How could I convert the $$\left(\frac{2}{3}\right)^{n}$$ term into $$\left(\frac{9}{4}\right)^{n}$$? Thanks.

• Try the inverse multiplication of what you did, you should get a good result. Specifically, multiply top and bottom of the original by $\left(\dfrac 23\right)^{2n+1}$. – abiessu Aug 14 at 2:58

A good general rule: divide numerator and denominator by the largest term: $$\frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}} =\frac{\frac{2}{3}}{\left(\frac{2}{3}\right)^{3n}+1}$$ and I think you should now be able to see what happens as $$n\to\infty$$.
$$\dfrac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}} =\dfrac1{\left(\frac{2}{3}\right)^{3n-1}+\frac{3}{2}}$$
Since $$\dfrac{2}{3} <1$$ , its limit will be zero.
Thus the answer is $$\dfrac{1}{0+\frac{3}{2}}=\dfrac{2}{3}$$