# About the double limit of a sequence

My book states that if the double limit $$\lim_{m,n}f(m,n)$$ exists and is equal to $$l$$ and if $$\lim_nf(m,n)$$ exists for all $$m$$, then $$\lim_m\lim_nf(m,n)$$ exists and is also equal to $$l$$. I was able to prove this result. However, the author gives the example : $$f(m,n)=\frac{1-(-1)^n}m$$ and says that $$\lim_nf(m,n)$$ does not exist and hence the double limit cannot exist. I don't see how he made that conclusion. The hypothesis requires first the existence of the limit $$\lim_nf(m,n)$$ before we can conclude anything.

Also, I think the double limit does exist and is equal to $$0$$, since $$|f(m,n)-0|≤\frac2m \forall m,n$$.

• I agree, the double limit and the limit for $m$ both exist and the limit for $n$ doesn't exist.
– user526015
Feb 13, 2019 at 12:46

It would be better, perhaps, to say that, while $$\lim_{m,n\to\infty}f(m,n)=0,\tag{1}$$ we have by contrast that $$\lim_{m\to\infty}\lim_{n\to\infty}f(m,n)$$ does not exist. That latter could be considered a double limit of a sort, but "iterated limit" would be more precise.
By your observation, we can, indeed, prove $$(1),$$ and further can prove that $$\lim_{m\to\infty}f(m,n)=0$$ for all $$n,$$ whence by the result you proved, we have $$\lim_{m,n\to\infty}f(m,n)=\lim_{n\to\infty}\lim_{m\to\infty}f(m,n).$$ This function $$f$$ therefore provides a good example of where a double-limit can be rewritten as an iterated limit in one way, but not the other.