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?)

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 $m_n:=\lceil\sqrt n\rceil=o(n)$,
$$x_{n,m_n}=\exp\left(\log n\cdot\log\left(1-\frac1{\lceil\sqrt n\rceil}\right)\!\right)=\exp\left(-\frac{\log n}{\sqrt n}+o\!\left(\frac{\log n}{\sqrt n}\right)\!\right)\xrightarrow[n\to\infty]{}1.$$
