# Limit $\lim\limits_{n \rightarrow \infty} \left(\lim\limits_{k \rightarrow \infty} \left(\frac{1}{1+2^{n-k}}\right) \right)$

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:

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

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

• Limes?${}{}{}{}{}{}{}$ – Omnomnomnom Oct 13 '15 at 14:44
• @Omnomnomnom limeslimits.wordpress.com – imranfat Oct 13 '15 at 14:45
• Do you mean lime trees or lime citrus? – Bernard Oct 13 '15 at 14:48
• @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 – imranfat Oct 13 '15 at 14:49
• Im sorry english is not my native language, of course lime trees! Seriously, I mean limes as the limit of a sequence – gammaALpha Oct 13 '15 at 14:50

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 $$\lim_{n\rightarrow\infty}1=1$$