# limit of a double exponential

How would I evaluate

$$\lim_{n\to\infty} \frac{A^{2^{n+1}} + \frac{1}{2}}{\left({A^{2^{n}} + \frac{1}{2}}\right)^2}$$

where A is a constant greater than 1?

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$$\lim_{n\to\infty} \frac{A^{2^{n+1}} + \frac{1}{2}}{\left({A^{2^{n}} + \frac{1}{2}}\right)^2}=1$$ –  Ethan Jan 20 '13 at 3:33

## 2 Answers

$$\lim_{n\to\infty} \frac{A^{2^{n+1}} + \frac{1}{2}}{\left({A^{2^{n}} + \frac{1}{2}}\right)^2}=\lim_{n\to\infty} \frac{1 + \frac{1}{2A^{2^{n+1}}}}{\frac{1}{A^{2^{n+1}}}\left({A^{2^{n}} + \frac{1}{2}}\right)^2}=\lim_{n\to\infty} \frac{1 + \frac{1}{2A^{2^{n+1}}}}{(\frac{1}{A^{2^{n}}})^2\left({A^{2^{n}} + \frac{1}{2}}\right)^2}=\lim_{n\to\infty} \frac{1 + \frac{1}{2A^{2^{n+1}}}}{\left({1 + \frac{1}{2A^{2^n}}}\right)^2}=1$$

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Hint: Divide the numerator and the denominator by $A^{2^{n+1}}$.

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