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This question is inspired from another much easier problem I was trying to solve which I tried to generalize. The question is essentially as follows (assuming all the limits exist)

If $a\in \mathbb R\cup \{\infty\}$, and $f$ and $g$ are two functions such that $$\lim_{x\to a}\left|f(x)-g(x)\right|=0$$ and $\varphi$ and $\tau$ are functions such that $$\lim_{x\to \lim_{t\to a}f(t)}\left|\varphi (x)-\tau (x)\right|=0$$ then what is the least amount of conditions necessary to impose so that $$\lim_{x\to a}\left|\varphi\circ f(x)-\tau \circ g(x)\right|=0$$ can be concluded?

What about the special case $f=\varphi$ and $g=\tau$?

In the original problem I had, I was in the special case, and dealing with $a=\infty$ which is why I was only considering those cases. I was under the impression that the continuity of $g$ and the extra condition that $$\lim_{x\to \infty} |f(x)|=\infty$$ would be enough. But, I soon arrived at a counterexample: construct $f$ and $g$ such that $$\lim_{x\to \infty} f(x)= \lim_{x\to \infty} g(x) = -\infty$$ but $$\lim_{x\to -\infty} f(x) = 1,\quad \lim_{x\to -\infty} g(x) = 2$$ which shattered my hopes.

Also, it seems strange that there is very little (easily available) literature on such questions about composition of functions. So, does anyone know of any such result that deals with such a problem?

Link to MO post: https://mathoverflow.net/q/451283/311366

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    $\begingroup$ This seems too general: we don't know anything about $\phi(f(x))$ when $x$ is close to $a$, because $f(x)$ doesn't have to be close to $a$, and we only have condition about $\phi$ for argument near $a$. $\endgroup$
    – mihaild
    Commented Jul 23, 2023 at 8:47
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    $\begingroup$ @mihaild if you think that to be the case, you are free to impose the extra conditions that $$\lim_{x\to a} f(x) = a$$ which is essentially what I also did in the second case that I explained. Then, the question would be, how more can you loosen it? $\endgroup$ Commented Jul 23, 2023 at 8:57
  • $\begingroup$ What was the original easier problem? It may be that this is not the best path to generalisation $\endgroup$
    – FShrike
    Commented Jul 23, 2023 at 10:09
  • $\begingroup$ @FShrike I had a function $f$, and I knew that $y=mx$ was its asymptote. I was trying to show that the asymptote of $f^{(n)}$ is $y=m^n x$ $\endgroup$ Commented Jul 23, 2023 at 11:24
  • $\begingroup$ @mihaild have a look at the edit $\endgroup$ Commented Jul 23, 2023 at 11:28

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