I have to calculate the following limit:

$A=\displaystyle \lim_{n \rightarrow \infty} \left(\displaystyle \lim_{k \rightarrow \infty} \left(\frac{1}{1+2^{n-k}}\right) \right) $

Where $\ n $ and $\ k $ are elements of the natural numbers.

Because we never did that in class, I am not sure as what I did is right. I got the following:

$\ 1 \leqslant A=\displaystyle \lim_{n \rightarrow \infty} (\displaystyle \lim_{k \rightarrow \infty} (\frac{1}{1+2^{n-k}}) ) \leqslant \displaystyle \lim_{n \rightarrow \infty} (\displaystyle \lim_{k \rightarrow \infty} (\frac{1}{2^{n-k}}) ) \leqslant \displaystyle \lim_{n \rightarrow \infty} (\displaystyle \lim_{k \rightarrow \infty} (\frac{2^{k}}{2^{n}}) ) = \frac{\displaystyle \lim_{k \rightarrow \infty}2^{k}}{\displaystyle \lim_{n \rightarrow \infty}2^{n}} $

Now its obvious that both limes are going to infinity, but how can I argue that they both like start at the same index $\ k $ and $\ n $ and so we could substitute both to the same variable and take the same limes so we would get the limit of 1 and then I could use the Sandwich theorem to tell that the limit of A is equal to 1.

Thank you

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    $\begingroup$ Limes?${}{}{}{}{}{}{}$ $\endgroup$ – Ben Grossmann Oct 13 '15 at 14:44
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    $\begingroup$ @Omnomnomnom limeslimits.wordpress.com $\endgroup$ – imranfat Oct 13 '15 at 14:45
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    $\begingroup$ Do you mean lime trees or lime citrus? $\endgroup$ – Bernard Oct 13 '15 at 14:48
  • $\begingroup$ @gammaALpha What worries me a bit is the order in which the limits for $k$ and $n$ are taken. That changes the outcome. Since the k-limit is inside the brackets, doe we have to consider that limit first? Or do we look at the limits simultaneously? In that case, the exponent of $2$ becomes indeterminate $\endgroup$ – imranfat Oct 13 '15 at 14:49
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    $\begingroup$ Im sorry english is not my native language, of course lime trees! Seriously, I mean limes as the limit of a sequence $\endgroup$ – gammaALpha Oct 13 '15 at 14:50

The order of the two limits is important. If you reverse the order you get a different answer. You need to do the inner one first then the outer one.

As the inner limit doesn't depend on $n$ you can rewrite it as: $$\lim_{n\rightarrow\infty}(2^{-n}\lim_{k\rightarrow\infty}\frac{1}{2^{-n}+2^{-k}})$$ The inner limit is then just $\frac{1}{2^{-n}}$ which would cancel giving you $$\lim_{n\rightarrow\infty}1=1$$

Or did you mean a limit where both are happening at the same time? If so you shouldn't use two different variables.

  • $\begingroup$ I see now that the order is important, and no not a limit where both are happening, this was just a thought. So do i have to do first the inner limit and then the outer? This means in this case the limit would be inifinty? $\endgroup$ – gammaALpha Oct 13 '15 at 14:53
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    $\begingroup$ Updated to show how to get the limit. $\endgroup$ – Ian Miller Oct 13 '15 at 15:00

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