Exactly when does a limit of a rational function exist?

Question

Let $f(x) = p(x)/q(x)$ be a rational function. Let q have a root at some $x = a$. It is clear that if $q(x)$ divides $p(x)$ then the limit of $f$ as $x$ goes to a exists. But is the converse true as well?

Answer 1

The converse is not necessarily true.

For example $$ \lim _{x\to 2} \frac {x^2-4}{x^2-5x+6} = -4 $$

But $(x^2-5x+6)$ does not divide $(x^2-4).$

Answer 2

If $q(x)$ divides $p(x)$, the quotient will be a polynomial, so the converse is also true.

In general instead, if $x=a$ is a zero for $q(x)$, i.e a pole for the rational function, and $f(a)$ (in the limit) is finite, then it means that $x=a$ is also a zero for $p(x)$ with multiplicity equal or greater that that of $p(x)$ and vice-versa, when you also consider multiplicity to be null.

Answer 3

The fact that $q(x)$ divides $p(x)$ cannot be derived if $$ \lim_{x\to a}\frac{p(x)}{q(x)} $$ exists, be $a$ a root of $q(x)$ or not.

Obviously, if $a$ is not a root of $q(x)$ the limit exists, by continuity.

If $a$ is a root of $q(x)$, then you can write $q(x)=(x-a)^mq_1(x)$, where $a$ is not a root of $q_1(x)$: just divide by $x-a$ as many times as you can.

In this case $m>0$ by assumption. You can also write $p(x)=(x-a)^np_1(x)$, with $a$ not a root of $p_1(x)$; here it is possible that $n=0$ (when $a$ is not a root of $p(x)$ to begin with. Then we have $$ \frac{p(x)}{q(x)}=\frac{(x-a)^n}{(x-a)^m}\frac{p_1(x)}{q_1(x)} $$ and, since $q_1(a)\ne0$, the limit $$ \lim_{x\to a}\frac{p_1(x)}{q_1(x)} $$ exists finite. So we just need to look at $$ \lim_{x\to a}\frac{(x-a)^n}{(x-a)^m} \tag{*} $$ This exists finite if and only if $n\ge m$, because otherwise it can be rewritten as $$ \lim_{x\to a}\frac{1}{(x-a)^{m-n}} $$ which does not exist if $m-n$ is odd (it is $\infty$ from the right and $-\infty$ from the left) and is $\infty$ if $m-n$ is even.

If $n=m$, then the limit ($\text{*}$) is $1$; if $n>m$, then the limit ($\text{*}$) is $0$.

Just play with some rational function to understand what happens: $$ \lim_{x\to 1}\frac{x^2+x-2}{x^2-1}=2 $$ but obviously $(x-1)(x+1)$ does not divide $(x-1)(x+2)$.

If you instead mean that the limit exists finite for all roots of $q(x)$, then the statement is false nonetheless. For instance $$ \lim_{x\to1}\frac{x^2-1}{x^3-x^2+x-1} $$ exists finite, and $1$ is the only (real) root of $x^3-x^2+x-1$.