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$\begingroup$

Just checking. $2^n$ ($n \to \infty$) tends to $\infty$.
+ $3^n$ also ($n \to \infty$) tends to $\infty$

so the sum gets me $\infty$.

Now $(\infty)^{1/\infty}$ : $(\infty)^0 = 1$

I see no other way. Theorem: $n^{\frac{1}{n}} = 1$ as $n \to \infty$.

Analogous: $(2^n - 3^n)^\frac{1}{n}$ = $(\infty - \infty)^0$ = 1 ???

Is it ok to suppose infinity on any integer $k^n$ as $n$ goes to infinity ?

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    $\infty^0$ is called an "indeterminate form", and your example is the reason why: when you have a limit that seems at first glance to have this form, you cannot tell what it will do without looking more closely.2012-05-07

2 Answers 2

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Before I answer your question: In general you really should be careful with takink limits seperately. You have probably seen proofs that taking limit behaves well under certain operations if the limit exists, i.e. is finite. When your limits are infinte there might be strange effects. Just look at the series for $e$: $ (1+1/n)^n $ For expressions like in your example the "sandwich lemma" is off big help. Just observe $3=(3^n)^{\frac 1n}\leq (2^n+3^n)^{\frac 1n}\leq (2\cdot 3^n)^{\frac 1n}=2^{\frac 1n}\cdot 3\to3$

For your "analogous" statement: You are aware that the expression in the brackets is negative for all $n\geq1$. Tking the $n^{th}$ root is not really well defined.

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    No, I would like to find out myself. But any books on the subject I have fail to clear out the basics for me. Any advise on a solid introduction to the subject, with good examples would be welcome.2012-05-10
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We can use following procedure: let $y= (2^n + 3^n)^{\frac{1}{n}}$. Then we can see that $\ln(y)=\frac{\ln(2^n+3^n)}{n}$ on the right side, limit is in the form infinity/infinity, so use L'hopitals rule, take derivatives, we would have $\frac{2^n\ln(2)+3^n\ln(3)}{2^n+3^n}$ it's limits is $ln(3)$, so finally we would have $y$ is equal to $e$ in power of $\ln(3)$, so finally $y=e^{ln(3)}=3$.

I think is it correct,if wrong tell me and i will change it

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    i have changed it,is not it correct now?2012-05-07