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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 - $25^n$.

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


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

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

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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
up vote 9 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 $\lim\limits_{n \rightarrow \infty} \frac{2^{2n+1}+3 \cdot5^{2n +1}}{2 \cdot 3^{2n-1} + 25^n} = 15$.


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 $-15$.

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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

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