Let $f(x)/p(x)$ and $g(x)/q(x)$ be rational functions in their simplest forms, with $\text{deg}(f)\neq\text{deg}(g)$ and $\text{deg}(p)\neq\text{deg}(q)$. Then $$\lim_{x\to\infty}\left[\frac{f(x)}{p(x)}+\frac{g(x)}{q(x)}\right] \,\text{exists}\implies \lim_{x\to\infty}\frac{f(x)+g(x)}{p(x)+q(x)}\,\text{exists}\,\,\,?$$
A seemingly true counterexample would be $$\lim_{x\to\infty}\left[\frac{x}{x^2}+\frac{1}{1-x^2}\right] =0\implies \lim_{x\to\infty}\frac{x+1}{1}=\infty$$ but $x/x^2=1/x$ is not in its simplest form and the denominators have the same degree.
We could also write $$\begin{align}\lim_{x\to\infty}\frac{f(x)+g(x)}{p(x)+q(x)}&=\lim_{x\to\infty}\left[\frac{f(x)}{p(x)+q(x)}+\frac{g(x)}{p(x)+q(x)}\right]\\&=\lim_{x\to\infty}\left[\frac{f/p}{1+q/p}+\frac{g/q}{1+p/q}\right]\end{align}$$ and this would prove the statement if $\lim_{x\to\infty} q/p\neq0,-1$ exists.
An example would be $\,\frac{x^2}{x-1} + \frac{-x}{1}\,$ which goes to $\,1\,$ but $\,\frac{x^2 - x}{x - 1 + 1} = x-1\,$ goes to $\,\infty\,$ as $\,x\to\infty.$
However, this example depends on one numerator having negative leading sign. In we restrict to polynomials with positive leading coefficient, then $\, f/p+g/q \to L < \infty \,$ does imply that $\,(f+g)/(p+q) \to M < \infty.\,$ As abbreviation, let $\,dp := \text{deg}(p)\,$ for any polynomial $\,p\,.$ Since $\, f/p+g/q \to L < \infty \,$ and both $\,f/q\,$ and $\,g/q\,$ must converge, this means $\,df \leq dp \,$ and $\,dg \leq dq.\,$ We are given $\,dp \neq dq\,$ so assume WLOG that $\,dp >dq.\,$ Now we are given $\,df \neq dg\,$ so there are two cases.
Case $1$: $\,df < dg.\,$ In $\,(f\!+\!g)/(p\!+\!q),\,$ we have $\,d(f\!+\!q) = dg \leq dq < dp = d(p\!+\!q).\,$ Now$\,(f\!+\!g)/(p\!+\!q) \! \to \!0 .\,$
Case $2$: $\, df > dg.\,$ We have $\,d(f\!+\!q) = df \leq dp = d(p\!+\!q).\,$ Now$\,(f\!+\!g)/(p\!+\!q) \! \to \!M <\infty .\,$