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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 $|f(m,n)-0|≤\frac2m \forall m,n$.

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

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This seems to be a case of ambiguous expression, but it could be mis-statement.

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.

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