There is sequence $a_{m,n}=\frac{m}{m+n}$ we calculate the following limits $$\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n} \qquad \lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}a_{m,n}$$

I find both of these limits to be $1,0$ respectively. But the limit of $$\lim_{m,n\rightarrow\infty}a_{m,n}=0.5$$ should be $0.5$ because we have a denominator that will be twice the numerator for very large but comparable values of $m,n$. What is the notion of limits in this situation?

Edit: Can the simultaneous limit be written like this

Since $m,n\rightarrow \infty \implies m \approx n\implies \lim_{m,n\rightarrow \infty}a_{m,n}=\lim_{n\rightarrow \infty}\frac{n}{n+n}=0.5$

  • 6
    $\begingroup$ There is no requirement that the values of $m$ and $n$ be comparable in $\lim_{m,n\to\infty}a_{m,n}$. The limit should hold no matter how I send $m$ and $n$ to infinity, whether as $m=n$, or $m=2n$, or $n=m^2$, etc. What you have computed instead is $\lim_{m\to\infty}a_{m,m}$. $\endgroup$
    – user856
    Feb 20 '18 at 7:12

For fixed $n$, we have $\lim_{m\rightarrow\infty}a_{m,n}=\lim_{m\rightarrow\infty}\dfrac{m}{m+n}=1$, so $\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n}=\lim_{n\rightarrow\infty}1=1$.

The limit $\lim_{m,n\rightarrow\infty}a_{m,n}$ does not exist. If it were, then $\lim_{m,n\rightarrow\infty}a_{m,n}=\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n}=\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}a_{m,n}$, but in this case they are not equal.

$|a_{m,n}-L|<\epsilon$ for $m,n\geq N$ is the formalised meaning of $\lim_{m,n\rightarrow\infty}a_{m,n}=L$, in which case such $m,n$ vary freely from $N$, neither of which bounds the other, so this is in some sense that $m,n$ need no be comparable, they are independent, as @Rahul, @Arthur have noted.

  • $\begingroup$ You need to fix the outer variable, in this case, fixing $n$ first, and look at the sequence $\{a_{m,n}\}_{m=1}^{\infty}$. $\endgroup$
    – user284331
    Feb 20 '18 at 6:59
  • $\begingroup$ Thank you again with another form :-)). $\endgroup$
    – Sebastiano
    Nov 30 '19 at 22:44

Why do you assume the denominator will be about twice the numerator? That is like saying that $m\approx n$,which need not be true. Sure, you could calculate what the limit is when $m=n$, and it will be $\tfrac12$, as you can iterate limits, or you could think that $n$ grows twice at fast as $m$. As long as $m\to \infty$, then will $n\to \infty$ too, but for $m=100$ it will be $n=200$; if $m=500$ then $n=1000$; if $m=2\times 10^8$, then $n=4\times 10^8$... and so on.

Do the math for each pair of numbers and you'll see that they always yield $\tfrac13$ as result, so it seems fine to say that as $m,n\to\infty$ keeping the relation $n=2m$ the limit is $\tfrac13$, which you could have calculated checking that $$\lim\limits_{n=2m} \frac m {m+n}=\lim_m \frac m {m+2m}=\lim_m \frac m {3m}=\lim_m \frac 13=\frac13.$$

Even more: suppose that $n$ grows much faster than $m$, although both tend to $\infty$, say $n=m^2$ (so that when $m$ reaches $100$, $n$ will be $10 000$, for instance): what happens now? It would be reasonable to say that $n$ wins in the long run, so the denominator will be much bigger than the numerator and the limit will be zero: again.

Anyway, that's just an intuition, you can see that $$\lim\limits_{n=m^2} \frac m {m+n}=\lim_m \frac m {m+m^2}=\lim_m \frac 1 {1+m}=0.$$

The thing here is that by fixing a relation between $m$ and $n$ you are coming back to a limit with only one variable/index, and then there wouldn't be anything new. Although these limits with restrictions are useful many times, for doble sequences one wish to define a concept of limit that doesn't have to consider any particular path or relation. Well, in this case, that won't be possible, since changing the relation the limit also changes it's value.

But there are situations in which the double or simultaneous limit is well defined (and I haven't dealt with the definition of limit, just a few intuitive arguments). Consider $$\lim_{m,n} \frac {2m} {m^2+nm}=\lim_{m,n} \frac 2 {m+n}=\cdots$$ Since the numerator is constant, and the denominator will tend to infinity —no matter the 'relative velocities' of $m$ and $n$—, then the limit has to be zero (and it is what we would be able to prove using the definition of limit). You can even check that using the same relations as before, and iterating limits, all give also zero.

(IMPORTANT: But that last idea doesn't work backwards, not fully at least: you would never have enough examples of relations between $m$ and $n$ that make the limit go to zero, to be sure that the double limit is zero; there are infinite options and there could always be one you haven't tried and gives a different answer; and if that happened to be the case, what you can assert is that the double/simultaneous limit does not exist).

  • $\begingroup$ Thanks a lot for taking the time to write this. Much appreciated. $\endgroup$
    – Sonal_sqrt
    Feb 20 '18 at 7:53
  • $\begingroup$ You're welcome. I'm glad that you find it useful! $\endgroup$ Feb 20 '18 at 18:17

Simultaneous limit does not mean that $m$ and $n$ grow equally fast. It means that they grow independently. If, as they grow independently, there isn't one single limit that the expression tends to, then the simultaneous limit doesn't exist.

  • $\begingroup$ How do you formalize the notion of growing independently? $\endgroup$
    – Alex Ortiz
    Feb 20 '18 at 7:11
  • 1
    $\begingroup$ $|a_{m,n}-L|<\epsilon$ for $m,n\geq N$, so $m$ and $n$ can be freely vary from $N$, neither of which bounds the other. $\endgroup$
    – user284331
    Feb 20 '18 at 7:13
  • 2
    $\begingroup$ @AOrtiz The standard way is to say "$\lim_{m,n\to\infty}a_{m,n}=L$ if, for any $\epsilon>0$, there is an $N\in \Bbb N$ such that for any $m,n>N$ we have $|a_{m,n}-L|<\epsilon$" This might look like they're increasing dependently, but notice that $N$ is just a lower bound. As long as they're both larger than $N$, one of them could be massively larger than the other, or they could be equal, or anything in-between. $\endgroup$
    – Arthur
    Feb 20 '18 at 7:16
  • $\begingroup$ In a sense, $a_{m,n}$ has all real numbers between $0$ and $1$ as partial limits. You can look at $a_{m_i,n_i}$ s.t. for every $i$ either $n_{i+1}=n_i+1 \land m_{i+1}=m_i$ or $n_{i+1}=n_i \land m_{i+1}=m_i+1$, choosing which one to increment according to whether you're over or under your desired value. $\endgroup$
    – Itai
    Feb 20 '18 at 8:17

"Simultaneous limits", as you call them, $\lim_{m,n\to\infty}$ do not mean that $n=m$ on all evaluations. They mean that $n,m$ can take all pairs of values that tend to infinity, and the argument is $a_{mn}$. They are 2D limits.

In iterated limits, you deal twice with 1D limits, and


where the $b_m$ need not belong to the set of $a_{mn}$.

These limits may exist/not exist independently and take different values.


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