The list of techniques to evaluate a limit that you have given above covers most of the non-elementary limit problems. Below I have given a few results that will be helpful in evaluating certain limit problems$:$
$ \bullet \ $ Assume that $f:(-a,a) \setminus \{0\} \rightarrow \Re \ $. Then $\lim_{x\to0}f(x)=l \ $ if and only if $ \ \lim_{x\to0}f(\sin(x))=l. $
$\bullet \ $ Assume that $f:(-a,a) \setminus \{0\} \rightarrow \Re \ $ . If $ \lim_{x\to0}f(x)=l \ $ then $ \lim_{x\to0}f(\left \vert {x} \vert \right) = l. $
$\bullet \ $ $ lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x = e$
$\bullet$ $ lim_{x\to 0} (1+x)^\frac{1}{x} = e $
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.
Best Answer
The limits are not equivalent. There are some mistakes in the referred question. Note, the title in the referred question states \begin{align*} \lim_{x\to\infty}\left(\left(x^\color{blue}{5}+x^\color{blue}{4}\right)^{\frac{1}{6}}-\left(x^\color{blue}{5}-x^\color{blue}{4}\right)^{\frac{1}{6}}\right) \end{align*} whereas the body of the referred question states \begin{align*} \lim_{x\to\infty}\left(\left(x^\color{blue}{6}+x^\color{blue}{5}\right)^{\frac{1}{6}}-\left(x^\color{blue}{6}-x^\color{blue}{5}\right)^{\frac{1}{6}}\right) \end{align*}
Taking the formula from the title we obtain \begin{align*} \lim_{x\to\infty}&\left(\left(x^5+x^4\right)^{\frac{1}{6}}-\left(x^5-x^4\right)^{\frac{1}{6}}\right)\\ &=\lim_{x\to\infty}\frac{\left(\sqrt[6]{x^5+x^4}-\sqrt[6]{x^5-x^4}\right)\left(\sqrt[6]{x^5+x^4}+\sqrt[6]{x^5-x^4}\right)} {\sqrt[6]{x^5+x^4}+\sqrt[6]{x^5-x^4}}\\ &=\lim_{x\to\infty}\frac{\sqrt[3]{x^5+x^4}-\sqrt[3]{x^5-x^4}} {x^{\frac{5}{6}}\left(\sqrt[6]{1+\frac{1}{x}}+\sqrt[6]{1-\frac{1}{x}}\right)}\\ &=\lim_{x\to\infty}\frac{x^{\frac{5}{6}}\left(\sqrt[3]{1+\frac{1}{x}}-\sqrt[3]{1-\frac{1}{x}}\right)} {\sqrt[6]{1+\frac{1}{x}}+\sqrt[6]{1-\frac{1}{x}}}\\ \end{align*}
Hint: One answer in the referred question mixes up expressions $x^5-x^4$ with $x^6-x^5$ and similar things.
We recall \begin{align*} a^6-b^6&=\left(a-b\right)\left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)\\ &=(a-b)\sum_{j=0}^5a^{5-j}b^{j}\tag{2} \end{align*} Letting \begin{align*} a&=\left(x^5+x^4\right)^{\frac{1}{6}}=x^{\frac{5}{6}}\left(1+\frac{1}{x}\right)^{\frac{1}{6}}\\ b&=\left(x^5-x^4\right)^{\frac{1}{6}}=x^{\frac{5}{6}}\left(1-\frac{1}{x}\right)^{\frac{1}{6}}\\ \end{align*}
Hint: The following is valid \begin{align*} \color{blue}{\lim_{x\to\infty}\left(\left(x^6+x^5\right)^{\frac{1}{6}}-\left(x^6-x^5\right)^{\frac{1}{6}}\right)=\frac{1}{3}} \end{align*} Some nice answers can be found here.