# Thread: limit for a sequence...

1. ## limit for a sequence...

Hey all,

Having trouble trying to find the limit for the following sequence
$\frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$ as the sequence goes from $n = 1$ to $\infty$

Thanks for all the help....

2. ## Re: limit for a sequence...

$\lim_{n \rightarrow \infty}\frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$ aquires the indeterminate form, $\frac{\infty}{\infty}$. So I would suggest the use of L' Hopital's rule.

Also check what wolfram alpha has to say:chick here

Thanks :-)

4. ## Re: limit for a sequence...

Not to dissuade you from trying this, but I don't think L'Hopital's will help here. First off, it is of indeterminate form $\frac 0 0$. Furthermore, since these are all exponential functions, they will all reappear when applying L'Hopital's rule, and will only be multiplied by constants, so you will still have an indeterminate form. WolframAlpha doesn't show the steps, unfortunately.

5. ## Re: limit for a sequence...

Yes, I am sorry. It is of the indeterminate form $\frac{0}{0}$.

$\lim_{n \rightarrow \infty}\frac{(\frac{1}{2}) ^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$

$=\lim_{n \rightarrow \infty}\frac{2^{-2n-1}-3^{1-3n}}{3^{1-2n}+4^{-n}}$

Tried L' Hopital's. It is also of no use here.

Wolfram Alpha shows (1/2) but there seems no way to solve it.

6. ## Re: limit for a sequence...

Originally Posted by Oiler
Hey all,

Having trouble trying to find the limit for the following sequence
$\frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$ as the sequence goes from $n = 1$ to [TEX]\infty[TEX]
Note
$\frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}=$ $\frac{(\frac{1}{4})^{n}(\frac{1}{2})-(\frac{1}{27})^{n}(3)}{(\frac{1}{4})^{n}+(\frac{1} {9})^{n}(3)}$

Dividing by $\left(\tfrac{1}{4}\right)^n$ it is clear that the limit is $\frac{1}{2}$.