Does $x_{n, m_n} \overset{n \to \infty}{\to} x$ hold, if $m_n = o(n)$ and $x_{n,m} \overset{n \to \infty}{\to} x_m \overset{m \to \infty}{\to} x$? Suppose we have $(x_{n,m}) \subset \mathbb R$ and $(x_m)\subset \mathbb R$ and $x \in \mathbb R$ such that

*

*For all $ m \in \mathbb N$ we have $x_{n,m} \overset{n \to \infty}{\longrightarrow} x_m \,$; and

*$x_m \overset{m \to \infty}{\longrightarrow} x$.

Let $(m_n)_n \subseteq \mathbb N$ be an non-decreasing sequence of indices such that $m_n \overset{n \to \infty}{\longrightarrow} \infty$ and $m_n = o(n)$.
Does it then hold that $x_{n, m_n} \overset{n \to \infty}{\longrightarrow} x$?

My intuition (not rigorous):
I think this could be correct, since $m_n = o(n)$ means that $m_n$ is kind of "standing still" compared to $n$, so $\lim_{n \to \infty} x_{n, m_n} = \lim_{m \to \infty} \lim_{n \to \infty} x_{n, m}$.

Thoughts/Attempt:
Using the triangle inequality, we get
$$
|x_{n,m_n} - x| \leq |x_{n, m_n} - x_{m_n}| + |x_{m_n} - x|.
$$
We know that $|x_{m_n} - x| \overset{n \to \infty}{\longrightarrow} 0$.
Therefore, it would suffice to show $|x_{n, m_n} - x_{m_n}| \overset{n \to \infty}{\longrightarrow} 0$.
I'm hoping that $m_n = o(n)$ will guarantee than this converges to zero. But I don't know if that's the case.
Possible problem: What if $(|x_{n,m} - x_m|)_n$ converges increasingly slowly as $m$ gets larger? Could the $|x_{n,m_n} - x_{m_n}|$ then not converge to zero?
Note: For my purposes at the moment, I just need this for numbers in $[0,1]$. So I could replace $\mathbb R$ above with $[0,1]$. (Maybe this works, but only on compact sets?)
 A: Let $$x_{n,m}:=\left(1-\frac1m\right)^{\log n}.$$
Then $x_{n,m}\xrightarrow[n\to\infty]{}0=:x_m$, and $x_m\xrightarrow[m\to\infty]{}0=:x.$
However, for $m_n:=\lceil\sqrt n\rceil=o(n)$,
$$x_{n,m_n}=\exp\left(\log n\cdot\log\left(1-\frac1{\lceil\sqrt n\rceil}\right)\!\right)=\exp\left(-\frac{\log n}{\sqrt n}+o\!\left(\frac{\log n}{\sqrt n}\right)\!\right)\xrightarrow[n\to\infty]{}1.$$
A: The little-o assumption doesn't change much. The convergence can be even slower than the growth of $m_n$ (whatever $m_n$ is!).
To see this, first consider the sequence
$$
y_{n,m} = 
\begin{cases}
0 & \text{if } n \le m, \\
1 & \text{if } n > m.
\end{cases}
$$
Clearly, $y_{n,m} \xrightarrow{n \to \infty} 1$ for each $m$, but the subsequence $y_{n,n}$ is constantly $0$ and doesn't converge to $1$. This corresponds to $m_n = n$, which is excluded in your question.
So let $m_n$ be any non-decreasing sequence such that $m_n \xrightarrow{n \to \infty} \infty$ and $\frac{m_n}{n} \xrightarrow{n \to \infty} 0$ (although the latter won't play any role). We will construct a counterexample $x_{n,m}$ as follows. First, define the sequence
$$
t_m := \max \{ n : m_n \le m \}.
$$
One should check from the definition that

*

*for each $m$, $\{ n : m_n \le m \}$ is a bounded subset $\mathbb{N}$ and thus has a maximum,

*$t_{m_n} \ge n$ for each $n$,

*$t_m \xrightarrow{m \to \infty} \infty$.

Now we define:
$$
x_{n,m} := y_{n,t_m}.
$$
Finally, it's easy to convince yourself that
$$
x_{n,m_n} = y_{n,t_{m_n}} = 0 \not\to 1,
$$
thanks to $n \le t_{m_n}$.

The construction is probably clearer if we distinguish between $n$ and $m$. Let's use the notation
$$
y^{(m)} = (y_{1,m},y_{2,m},y_{3,m},\ldots),
$$
so that we have a sequence $y^{(m)}$ (and each term is a sequence itself). Then the sequence $x^{(m)}$ is chosen from $y^{(m)}$ as a subsequence: $x^{(m)} = y^{(t_m)}$. It's not exactly a subsequence, as $t_m$ doesn't have to be strictly increasing, but in principle $t_m$ grows fast enough to compensate for the slowness of $m$. This last fact is encapsulated in the condition $t_{m_n} \ge n$.
