2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Try the inverse multiplication of what you did, you should get a good result. Specifically, multiply top and bottom of the original by $\left(\dfrac 23\right)^{2n+1}$. $\endgroup$ – abiessu Aug 14 at 2:58
5
$\begingroup$

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 $n\to\infty$.

Good luck!

$\endgroup$
  • $\begingroup$ Thank you very much $\endgroup$ – squenshl Aug 14 at 5:35
2
$\begingroup$

$$\dfrac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}} =\dfrac1{\left(\frac{2}{3}\right)^{3n-1}+\frac{3}{2}}$$

Since $\dfrac{2}{3} <1$ , its limit will be zero.

Thus the answer is $\dfrac{1}{0+\frac{3}{2}}=\dfrac{2}{3}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.