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
Answer
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.
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