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What is \(\displaystyle\lim_{n\rightarrow\infty}\dfrac{A}{B},\) where \(A=n^{(n-1)^{(n-2)^{\ldots^{3^2}}}}\) and \(B=2^{3^{4^{\ldots^{(n-1)^n}}}}\)? I suspect it is \(0,\) but I don't have any idea how I would go about proving this. A little help?

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Those terms should be defined as \( A_n\) and \(B_n\).

Hint: How does \( A_n \) and \( A_{n-1} \) relate? What about \( B_n \) and \( B _{n-1} \)?

Which test does that suggest we apply?

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While it is obvious that \(A_n=n^{A_{n-1}},\) I can't come up with a mathematical relartion for \(B_{n-1}\) and \(B_n,\) since \(B_n\neq B_{n-1}^n.\)

The way I look at it, the limit of \(\frac{A}{B}\) would be in the form \(\frac{\inf}{\inf}\),

Certainly, but the question would be which sequence (see Calvin Lin's comment) grows faster.

Oh, ok, thanks

Wait, I think im pretty close.

I have a question, though. If the B grows faster, than the equation would near zero. If A grows faster, the equation would near infinity... right?

Am I missing something right now?

suppose the value of pie =0.31825. calculate the area of circle of radius= 5 equal to 78.55 . please share the answer ok.

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## Comments

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TopNewestThose terms should be defined as \( A_n\) and \(B_n\).

Hint:How does \( A_n \) and \( A_{n-1} \) relate? What about \( B_n \) and \( B _{n-1} \)?Which test does that suggest we apply?

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While it is obvious that \(A_n=n^{A_{n-1}},\) I can't come up with a mathematical relartion for \(B_{n-1}\) and \(B_n,\) since \(B_n\neq B_{n-1}^n.\)

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The way I look at it, the limit of \(\frac{A}{B}\) would be in the form \(\frac{\inf}{\inf}\),

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Certainly, but the question would be which sequence (see Calvin Lin's comment) grows faster.

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Oh, ok, thanks

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Wait, I think im pretty close.

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I have a question, though. If the B grows faster, than the equation would near zero. If A grows faster, the equation would near infinity... right?

Am I missing something right now?

Log in to reply

suppose the value of pie =0.31825. calculate the area of circle of radius= 5 equal to 78.55 . please share the answer ok.

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