I need to calculate the limit of (1/3^(n+2)-1/5^(n+2))/(1/3^(n+1)-1/5^(n+1)) where n tends to infinity, i know the answer is 1/3 but I don't understand how you get that.

Can anyone explain? I will be grateful for any help.

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- Apr 10th 2011, 04:48 PMjoesmithcalculating a limit
I need to calculate the limit of (1/3^(n+2)-1/5^(n+2))/(1/3^(n+1)-1/5^(n+1)) where n tends to infinity, i know the answer is 1/3 but I don't understand how you get that.

Can anyone explain? I will be grateful for any help. - Apr 10th 2011, 05:17 PMtopsquark
Consider

$\displaystyle \displaystyle \frac{ 3^{-(n + 2)} - 5^{-(n + 2)}}{ 3^{-(n + 1)} - 5^{-(n + 1)}}$

$\displaystyle \displaystyle = \frac{ \frac{1}{9} \cdot 3^{-n} - \frac{1}{25} \cdot 5^{-n}}{ \frac{1}{3} \cdot 3^{-n} - \frac{1}{5} \cdot 5^{-n}}$

Now multiply top and bottom by $\displaystyle 3^n$ and see what happens.

-Dan - Apr 10th 2011, 05:37 PMjoesmith
Thanks i've just figured it out