Exercise 2.3.13

Question

[Iterated Limits]
  Given a doubly indexed array $a_{m n}$ where $m, n \in \mathbf{N}$, what should $\lim _{m, n ightarrow \infty} a_{m n}$ represent?
  \begin{enumerate}[label=(\alph*)] 
  \item Let $a_{m n}=m /(m+n)$ and compute the iterated limits
  $$
  \lim _{n ightarrow \infty}\left(\lim _{m ightarrow \infty} a_{m n}ight) \quad 	ext { and } \lim _{m ightarrow \infty}\left(\lim _{n ightarrow \infty} a_{m n}ight)
  $$
  Define $\lim _{m, n ightarrow \infty} a_{m n}=a$ to mean that for all $\epsilon>0$ there exists an $N \in \mathbf{N}$ such that if both $m, n \geq N$, then $\left|a_{m n}-aight|<\epsilon$
  \item Let $a_{m n}=1 /(m+n)$. Does $\lim _{m, n ightarrow \infty} a_{m n}$ exist in this case? Do the two iterated limits exist? How do these three values compare? Answer these same questions for $a_{m n}=m n /\left(m^{2}+n^{2}ight)$
  \item Produce an example where $\lim _{m, n ightarrow \infty} a_{m n}$ exists but where neither iterated limit can be computed.
  \item Assume $\lim _{m, n ightarrow \infty} a_{m n}=a$, and assume that for each fixed $m \in \mathbf{N}$, $\lim _{n ightarrow \infty}\left(a_{m n}ight) = b_{m}$. Show $\lim _{m ightarrow \infty} b_{m}=a$
  \item Prove that if $\lim _{m, n ightarrow \infty} a_{m n}$ exists and the iterated limits both exist, then all three limits must be equal.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item $$ \lim_{n o \infty}\left(\lim_{m o \infty} \frac{m}{m+n} ight) = 1, ext{ and } \lim_{m o \infty}\left(\lim_{n o \infty} \frac{m}{m+n} ight) = 0 $$ \item For $a_{mn} = 1/(m+n)$ all three limits are zero. For $a_{mn} = mn/(m^2 + n^2)$ iterated limits are zero, and $\lim_{m,n o \infty} a_{mn}$ does not exist since for $m,n \ge N$ setting $m=n$ gives $$ \frac{n^2}{n^2 + n^2} = \frac{1}{2} $$ Which cannot be made smaller then $\epsilon = 1/2$. The reason you would think to set $m=n$ is in trying to maximize $mn/(m^2 + n^2)$ notice if $m>n$ then $mn>n^2$ so we are adding more to the numerator then the denominator, hence the ratio is increasing. And if $mm$, $a_{mn}$ will oscillate between $1/m$ and $-1/m$, and thus $\lim_{n o \infty} a_{mn}$ does not exist. Similar reasoning shows that for a fixed $n$, $\lim_{m o \infty} a_{mn}$ does not exist either. But clearly $\lim_{m, n o \infty} a_{mn} = 0$. \item Choose $\epsilon > 0$ and let $0 < \epsilon' < \epsilon$. We need to find $N$ so that $|b_m - a| < \epsilon$ for all $m > N$. Set $N$ such that $|a_{mn} - a| < \epsilon'$ when $n,m \ge N$. Then fix $m \ge N$, I will show $|b_m-a|<\epsilon$. apply the triangle inequality to get $$ |b_m - a| \le |b_m - a_{mn}| + |a_{mn} - a|\quad\forall n \in \mathbf{N} $$ This inequality is true for all $n$, we will pick $n$ to make it strict enough to complete the proof. Set $n \ge \max\{N, N_m\}$ where $N_m$ (dependent on $m$) is big enough that $|b_m - a_{mn}| < \epsilon-\epsilon'$. We also have $|a_{mn} - a| < \epsilon'$ since $m \ge N$ and $n \ge N$. So finally $$ |b_m - a| \le |b_m - a_{mn}| + |a_{mn} - a| < (\epsilon-\epsilon') + \epsilon' = \epsilon $$ And we are done. The key is that we can make $|b_{m} - a_{mn}|$ as small as we want \emph{independent of m}, so we take the limit as $n o \infty$ to show $|b_m - a| \le |a_{mn} - a|$. % TODO: Replace proof by taking limit of both sides? \item Let $b_m = \lim_{n o \infty} (a_{mn})$, $c_n = \lim_{m o \infty} (a_{mn})$, and $a = \lim_{m,n o\infty} (a_{mn})$. In (d) we showed $(b_m) o a$; a similar argument shows $(c_n) o a$. Thus all three limits are equal to $a$. \end{enumerate}

Exercise 2.3.13

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Exercise 2.3.13

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