Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would one calculate the following limit with basic pre calculus knowledge? I would like to see the process of dividing this expression with the highest power - $25^n$.

$$\lim_{n \to \infty}\left({\frac{2^{2n+1}+3\cdot 5^{2n+1}}{2\cdot3^{2n-1}+25^{n}}}\right)$$

EDIT:

If this expression would have a $(-1)^{n}$ in it, what would the limit be.

$$\lim_{n \to \infty}\left({\frac{2^{2n+1}+3\cdot 5^{2n+1}}{2\cdot3^{2n-1}+(-1)^{n}25^{n}}}\right)$$

share|improve this question
2  
Nice of you to come back, but do you have anything to say about the answer Matthias left here two days ago? –  Gerry Myerson Nov 29 '11 at 22:24
add comment

1 Answer

up vote 8 down vote accepted

You have

$\begin{align} \frac{2^{2n+1}+3 \cdot5^{2n +1}}{2 \cdot 3^{2n-1} + 25^n} &= \frac{2 \cdot2^{2n}+15 \cdot 5^{2n}}{\frac{2}{3} \cdot 3^{2n} + 5^{2n}} \\ &= \frac{2 \cdot (\frac{2}{5})^{2n}+15}{\frac{2}{3} \cdot (\frac{3}{5})^{2n} + 1}\\ \end{align}$

and so you get $\lim\limits_{n \rightarrow \infty} \frac{2^{2n+1}+3 \cdot5^{2n +1}}{2 \cdot 3^{2n-1} + 25^n} = 15$.

Edit:

This $(-1)^n$ makes your sequence divergent. If you let $a_n = \frac{2^{2n+1}+3 \cdot5^{2n +1}}{2 \cdot 3^{2n-1} + (-1)^n 25^n} $ you can look at the subsequences $(a_{2n})_{n\in \mathbb{N}}$ and $(a_{2n +1 })_{n\in \mathbb{N}}$. A computation like the one above shows $$ a_{2n} = \frac{2 \cdot (\frac{2}{5})^{4n}+15}{\frac{2}{3} \cdot (\frac{3}{5})^{4n} + 1}$$ converging to $15$ and $$ a_{2n+1} = \frac{2 \cdot (\frac{2}{5})^{4n+2}+15}{\frac{2}{3} \cdot (\frac{3}{5})^{4n+2} - 1}$$ converging to $-15$.

share|improve this answer
    
Matt (not Matthias!), would you care to accept this answer, or were you looking for something different? –  The Chaz 2.0 Feb 26 '12 at 14:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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