I am trying to find to generalize the limit that involves all rational functions such as $\sum_{n=0}^{l}\frac{{a}_{n}{x}^{n}}{{b}_{n}{x}^{n}}$. I believe the best way of generalizing all of them is using a limit expansion at infinity.
For example if we take $$\left({\left(\frac{x^2+5}{x+5}\right)}^{1.9}+{\left({x^2-5x+25}\right)}^{1.1}\right)^{23/22}-\left({\left({x-5}\right)}^{1.9}+{\left({x^2-5x+25}\right)}^{1.1}\right)^{23/22}$$
I think one way is to take the series expansion of this part of the limit and apply it to the second part..
$$\left({\left(\frac{x^2+5}{x+5}\right)}^{1.9}+{\left({x^2-5x+25}\right)}^{1.1}\right)^{23/22}$$
But that seems difficult so instead I tried to divide ${\left({\left(\frac{1}{x^2-5x+25}\right)}^{1.1}\right)}^{23/22}$
$$\frac{\left({\left(\frac{x^2+5}{x+5}\right)}^{1.9}\left(\frac{1}{{x^2-5x+25}}\right)^{1.1}+1\right)^{23/22}-\left({\left({x-5}\right)}^{1.9}{\left(\frac{1}{x^2-5x+25}\right)}^{1.1}+1\right)^{23/22}}{{\left({\left(\frac{1}{x^2-5x+25}\right)}^{1.1}\right)}^{23/22}}$$
Then substituting $a={\left(\frac{x^2+5}{x+5}\right)}^{1.9}\left(\frac{1}{{x^2-5x+25}}\right)^{1.1}$ and $b={\left(x-5\right)}^{1.9}\left(\frac{1}{{x^2-5x+25}}\right)^{1.1}$
$$\frac{\left(a+1\right)^{23/22}-\left(b+1\right)^{23/22}}{{\left({\left(\frac{1}{x^2-5x+25}\right)}^{1.1}\right)}^{23/22}}$$
Then I took the laurent series of $\left(a+1\right)^{23/22}$ and $\left(a+1\right)^{23/22}$
But if I subsitiuted $a$ and $b$ back, (for the sake of shortcuts I denote them as $a(x)$ and $b(x)$).
$$\frac{\left({a(x)}^{23/22}+\frac{23{a(x)}^{1/22}}{22}+\frac{23}{968{a(x)}^{21/22}}\right)-\left({b(x)}^{23/22}+\frac{23{b(x)}^{1/22}}{22}+\frac{23}{968{b(x)}^{21/22}}\right)^{23/22}}{{\left({\left(\frac{1}{x^2-5x+25}\right)}^{1.1}\right)}^{23/22}}$$
But according to my graphing calculator and wolfram alpha this is not equal to... $\lim_{x\to\infty}\left({\left(\frac{x^2+5}{x+5}\right)}^{1.9}+{\left({x^2-5x+25}\right)}^{1.1}\right)^{23/22}-\left({\left({x-5}\right)}^{1.9}+{\left({x^2-5x+25}\right)}^{1.1}\right)^{23/22}$
Is it possible to continue with my method or is their another approach that can generalize all rational functions?
This is not an answer but it is too long for a comment.
Did gave a very elegant answer and an efficient solution to approach this problem which looks to me as a nightmare.
Just for peronal fun, I decided to look at the asymptotics, considering $$A=\left(\left(\frac{x^2+5}{x+5}\right)^{19/10}+\left(x^2-5 x+25\right)^{11/10}\right)^{23/22}$$ $$B=\left((x-5)^{19/10}+\left(x^2-5 x+25\right)^{11/10}\right)^{23/22}$$ and developed (funny work !) for infinite values of $x$. The fourteen first terms are identical in both expansions and, by the end, after so many cancellations, $$A-B=\frac{1311}{22}+\frac{1311}{484} \left(\frac{1}{x}\right)^{3/10}+O\left(\left(\frac{1}{x}\right)^{3/5}\right)$$