Understanding limits of rational functions at infinity

[Aramin]

Is there a way to distinguish between rational functions that have the same limit at both ends and those that don't? I think I might have answered my own question, but let's say I evaluate a rational function, and it turns out to be a coefficient ratio with no variables (3/2). Does that mean that function will have the same end behavior on both sides?

What is required to have a result of -3/2 at -infinity and 3/2 at +infinity? Does this result occur in rational functions?

 

[Simon Bridge]

Welcome to PF;

Is there a way to distinguish between rational functions that have the same limit at both ends and those that don't?

Yes - you evaluate the limits at each end.

I think I might have answered my own question, but let's say I evaluate a rational function, and it turns out to be a coefficient ratio with no variables (3/2). Does that mean that function will have the same end behavior on both sides?

"Evaluate a rational function" where? How? What does that phrase mean?

Being careful to say what you are talking about - explicitly - should help you here ;)

 

[Aramin]

Yes I definitely would evaluate the limits at each end. I was wondering if there are any rules or patterns in the rational function before being evaluated that might tip this behavior off.

Regarding the 2nd part of my post, I figured it out :)

 

[SteamKing]

At each end of what?

You know, there is a handy and clear mathematical notation which you can use to express which limits you are talking about.

 

[pasmith]

Is there a way to distinguish between rational functions that have the same limit at both ends and those that don't? I think I might have answered my own question, but let's say I evaluate a rational function, and it turns out to be a coefficient ratio with no variables (3/2). Does that mean that function will have the same end behavior on both sides?

What is required to have a result of -3/2 at -infinity and 3/2 at +infinity? Does this result occur in rational functions?

The limits as $|x| \to \infty$ of $r(x) = (\sum_{n=0}^N a_nx^n)/(\sum_{n=0}^M b_mx^m)$ with $a_N$ and $b_M$ both nonzero are the limits of $$\frac{a_N}{b_M}x^{n-m}.$$ Clearly this only tends to a finite limit if $n \leq m$. If $n < m$ then the both limits are zero. A non-zero limit is possible only in the case $n = m$, where $$\lim_{x \to \infty} r(x) = \lim_{x \to -\infty} r(x) = \frac{a_N}{b_M}.$$

 

[Blazejr]

Let's say that numenator is polynomial of degree m (with coefficient A before m-th degree term) and denominator is polynomial of degree n (with coefficient B), that is:
$$Q(x)=\frac{Ax^m+...}{Bx^n+...}$$

When you go to inifnity those higher order terms dominate. That is no matter what are the coefficients, if you go to big enough numbers highest order term will give number bigger (in absolute value) than other numbers. Therefore asymptotically our function behaves like $F(x)=\frac{A}{B} x^{m-n}$. This can be made precise by noting that limit of $Q(x)-F(x)$ at infinities is zero (if you know precise definition of limit, try to prove it as an exercise!).

We established that all information about our function is encoded in F. Well from that it is obvious that if n was greater than m then our function goes to zero at infinity. If m>n then it behaves like a polynomial of degree m-n; and those have limits of the same sign at both infinities if their degree is even. Ratio of A to B gives you information if it is positive at positive infinity.

Additionally if you already know about limits i advise you to look up what "big O" notation is, it is very helpful in determining asymptotic behaviour of functions.

 

[pwsnafu]

What is required to have a result of -3/2 at -infinity and 3/2 at +infinity? Does this result occur in rational functions?

Not for rational functions no. But it's easy to write functions with the properties you want.

First consider $\sqrt{x^2+1} = \sqrt{x^2(1+x^{-2})}= |x| \, \sqrt{1+x^{-2}}$. Now divide by $1+x$. We obtain
$\frac{|x| \, \sqrt{1+x^{-2}}}{x(1+x^{-1})}$.

If we assume $x>0$ then this is just $\frac{\sqrt{1+x^{-2}}}{1+x^{-1}}$ and so the limit to positive infinity is clearly +1.

If we assume $x<0$ then $|x|=-x$ and so we obtain $\frac{-\sqrt{1+x^{-2}}}{1+x^{-1}}$. The limit to negative infinity is -1.