Source : Thomas, Calculus ( Chapter on limits).
Context: The question asks for the behaviour of the function $f(x)= \frac {x+1} {x^2 +3}$as $x$ goes to $+\infty$ and $ - \infty$. The answer being that function $f$ admits of $y=0$ as asymptote, both to the right and to the left.
But, in order to explain the graph of $f$, I'd like to go a bit further than the original question and to determine its behaviour around $x=0$.
To do this I rewrite $f$ as :
$\large f(x)= \frac {x+1} {x^2 +3} = \frac { \frac {x+1} {3} } {\frac {x^2+3} {3}} = \frac {x+1} {3} \times \frac { 1} {\frac {x^2+3} {3}} = \bigg{(}\frac 13 x + \frac 13 \bigg{)}\times \frac { 1} { 1+ \frac {x^2} {3}}$
From this I would like to conclude that, since the second factor goes to $1$ as $x$ goes to $0$ , the first factor is preponderant , in such a way that, near $0$ , function $f$ behaves like $y= \frac 13 x + \frac 13$.
Desmos construction : https://www.desmos.com/calculator/heksb1tgoj
My question is : to which extent is this reasoning rigorous, as it stands? what could be answered to the following objections
(1) the first factor goes to $1/3$ as $x$ goes to zero, so why not conclude that function $f$ behaves overall like $y=(1/3) \times 1$ as $x$ approaches $0$ ? why applying the limiting process only to the second factor?
(2) if the second term were not $\frac { 1} { 1+ \frac {x^2} {3}}$ but $\frac { 1} { 1+ \frac {x} {3}}$, this second term would also go to $1$ ( as $x$ goes to $0$), but one would not want to conclude from this that ( in this hypothetical case) the global function behaves like $y= \frac 13 x + \frac 13$ near $x=0$, so why using ths reason in the first case, and not in the second?