3
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

When you're talking about limiting behavior, its not just about what the limit of the function is, but also the "speed" in which that function also approaches its limit

The limits of both functions $x$ and $x^2$ are $\infty$ as $x\rightarrow\infty$, however we would say that the limit $$ \lim_{x\rightarrow\infty}\frac{x}{x^2} \neq 1 $$ because $x^2$ gets so much bigger so much faster than $x$.

So for your first question, yes you can say that the function acts like $y=\frac{1}{3}$ around $x=0$ but also if you want to be more accurate you'd say it acts like $y=\frac{x+1}{3}$, and if you wanted to get even more accurate than that you'd add in the quadratic term.

As for your second question of why we can't say the say thing if you change that one term, its because then $x$ decays at the same rate so they impact each other on the same scale.

If you Taylor series expand around $0$ you get $$ f(x) = \frac{1}{3}+\frac{x}{3}-\frac{x^2}{9}+\mathcal{O}(x^3) $$

where you cut it off at is what your approximation is and you'll be able to find the error from there

$\endgroup$
2
  • $\begingroup$ I guess you aimed at writing: " we would not say that .... $=1 $ " ( instead of $\neq 1$). $\endgroup$ Commented Dec 5, 2022 at 15:14
  • 1
    $\begingroup$ Oh yes you're right that's my bad! $\endgroup$
    – wjmccann
    Commented Dec 5, 2022 at 15:55
1
$\begingroup$

To understand the behaviour at zero I would suggest using a linear approximation $$f(x)\approx f(x_0)(x-x_0)+x_0\quad \text{near $x_0$}$$ In this case this would indeed result in the same approximation as $f'(0)=\frac{1}{3}=f(0)$.

The same can also be achieved by using Taylor expansion around $0$: a term of the form $\frac{1}{1-x^a}=\sum x^{an},a\in\mathbb{N}$ would then contribuite in the linar term only if $a=1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .