Simple Poles In Rational Functions

Why the Sign on Each Side Matters

A rational function can become unbounded near an input where its denominator is zero. But knowing only that the denominator approaches zero does not tell us whether the function grows toward positive infinity or negative infinity. The sign on each side of that input determines the answer.

In this tutorial, we study the case in which the denominator has a zero of multiplicity one and the numerator is nonzero there. This situation is called a simple pole. The key is to separate the factor that changes sign at the input from the factors that remain nonzero nearby.

Central idea: Near a simple pole, the numerator and the nonvanishing part of the denominator keep their signs. The factor $x-c$ changes sign at $c$, so the function has opposite signs on the two sides and tends to opposite infinities.

What Counts as a Simple Pole?

Let $R(x)=\frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials. We focus on an input $c$ for which $Q(c)=0$, $Q'(c)\neq 0$, and $P(c)\neq 0$. The condition $Q'(c)\neq 0$ means that $Q$ has a zero of multiplicity one at $c$. The condition $P(c)\neq 0$ ensures that the numerator does not also vanish there.

By the factor theorem, $Q(x)=(x-c)G(x)$ for a polynomial $G$, and the simple-zero condition gives $G(c)\neq 0$. Thus, for $x\neq c$ near $c$,

R(x)=\frac{P(x)}{(x-c)G(x)}.

Because $P(c)$ and $G(c)$ are nonzero, their signs persist for all $x$ sufficiently close to $c$. Meanwhile, $x-c$ is negative to the left of $c$ and positive to the right. This is the source of the opposite one-sided signs.

Definition: In this setting, $R(x)=\frac{P(x)}{Q(x)}$ has a simple pole at $c$ if $Q(c)=0$, $Q'(c)\neq 0$, and $P(c)\neq 0$. Equivalently, $Q(x)=(x-c)G(x)$ with $G(c)\neq 0$, while $P(c)\neq 0$.

A Sign Rule for Both One-Sided Limits

The sign can be summarized by the nonzero number

A=\frac{P(c)}{G(c)}.

This number gives the sign of the part of the function that does not change sign near $c$. The remaining factor is $1/(x-c)$. If $A$ is positive, the function is negative on the left and positive on the right. If $A$ is negative, those signs are reversed.

Theorem: Suppose $R(x)=\frac{P(x)}{(x-c)G(x)}$, where $P$ and $G$ are continuous near $c$, and $P(c)\neq 0$ and $G(c)\neq 0$. Set $A=\frac{P(c)}{G(c)}$. Then $$ \lim_{x\to c^-}R(x)= \begin{cases} -\infty,& A>0,\\ +\infty,& A<0, \end{cases} \qquad \lim_{x\to c^+}R(x)= \begin{cases} +\infty,& A>0,\\ -\infty,& A<0. \end{cases} $$

Here $x\to c^-$ means that $x$ approaches $c$ through values less than $c$; $x\to c^+$ means it approaches through values greater than $c$. The theorem requires that the numerator and the remaining denominator factor stay nonzero at $c$. If either condition fails, this particular sign rule does not apply.

Why the Rule Works

We can justify the conclusion using continuity and signs. Since $P(c)\neq 0$, continuity ensures that $P(x)$ has the same sign as $P(c)$ when $x$ is sufficiently close to $c$. Similarly, $G(x)$ has the same sign as $G(c)$ nearby. Therefore the quotient $H(x)=P(x)/G(x)$ has the same sign as $A$ near $c$. In fact, by continuity, $H(x)$ approaches the nonzero value $A$.

For a direct magnitude argument, choose a positive number $\delta$ small enough that whenever $0<|x-c|<\delta$, $H(x)$ has the sign of $A$ and $|H(x)|>|A|/2$. This is possible because $H(x)\to A$ and $A\neq 0$. Then

|R(x)|=\frac{|H(x)|}{|x-c|} >\frac{|A|}{2|x-c|}.

As $x$ approaches $c$, the right-hand side grows without bound. The sign of $R(x)$ is the sign of $H(x)$ multiplied by the sign of $1/(x-c)$. On the left, $x-c<0$; on the right, $x-c>0$. This proves both the unbounded magnitude and the stated signs of the one-sided limits.

Factor out the vanishing term.
Write the denominator as $(x-c)G(x)$, with $G(c)\neq 0$.

Determine the stable sign.
Find the signs of $P(c)$ and $G(c)$, or equivalently of $A=P(c)/G(c)$.

Use the side of approach.
The factor $x-c$ is negative on the left and positive on the right; combine that with the stable sign.

State the limits separately.
Write the left-hand and right-hand limits with their signs of infinity.

Worked Examples

Worked Example: A Positive Stable Sign

Find the one-sided limits of $R(x)=\frac{2x+1}{x-3}$ as $x\to 3$.

The denominator is already factored: $x-3=(x-3)\cdot 1$. Thus $c=3$, $G(x)=1$, $P(3)=2(3)+1=7$, and $G(3)=1$. Therefore $A=7/1=7>0$.

For $x<3$, $x-3<0$, while the numerator is positive near $3$, so the quotient is negative and its magnitude grows without bound. For $x>3$, both numerator and denominator are positive, so the quotient is positive and grows without bound.

\lim_{x\to 3^-}\frac{2x+1}{x-3}=-\infty, \qquad \lim_{x\to 3^+}\frac{2x+1}{x-3}=+\infty.

Check with nearby exact inputs: at $x=2.9$, the numerator is $6.8$ and the denominator is $-0.1$, giving $R(2.9)=-68$. At $x=3.1$, the numerator is $7.2$ and the denominator is $0.1$, giving $R(3.1)=72$. These values have the predicted signs and large magnitudes.

Worked Example: A Denominator with Another Factor

Find both one-sided limits of $R(x)=\frac{x^2+1}{(x+2)(x+4)}$ at $c=-2$.

The vanishing factor is $x+2=x-(-2)$, and the other factor is $G(x)=x+4$. The numerator at $c=-2$ is $P(-2)=(-2)^2+1=5$, and $G(-2)=-2+4=2$. Hence $A=5/2>0$.

Near $-2$, the numerator and $x+4$ are positive. On the left of $-2$, $x+2<0$, so the denominator is negative and the quotient is negative. On the right, $x+2>0$, so the quotient is positive. Therefore

\lim_{x\to -2^-}\frac{x^2+1}{(x+2)(x+4)}=-\infty, \qquad \lim_{x\to -2^+}\frac{x^2+1}{(x+2)(x+4)}=+\infty.

For a numerical sign check, at $x=-2.1$ the numerator is $5.41$ and the denominator is $(-0.1)(1.9)=-0.19$, so the value is $-541/19$. At $x=-1.9$, the numerator is again $5.41$ and the denominator is $(0.1)(2.1)=0.21$, so the value is $541/21$. The left value is negative and the right value is positive, as required.

Worked Example: A Negative Stable Sign

Find the one-sided limits of $R(x)=\frac{x-4}{(x+2)(x+4)}$ as $x\to -2$.

Write the denominator as $(x-(-2))G(x)$, where $G(x)=x+4$. At $c=-2$, $P(-2)=-2-4=-6$, and $G(-2)=-2+4=2$. Thus $A=-6/2=-3<0$.

Near $-2$, the numerator is negative and $G(x)$ is positive. For $x<-2$, the factor $x+2$ is negative, making the full denominator negative. A negative numerator divided by a negative denominator is positive, so the left-hand limit is $+\infty$. For $x>-2$, the denominator is positive while the numerator is negative, so the right-hand limit is $-\infty$.

\lim_{x\to -2^-}\frac{x-4}{(x+2)(x+4)}=+\infty, \qquad \lim_{x\to -2^+}\frac{x-4}{(x+2)(x+4)}=-\infty.

Check at $x=-2.1$: the numerator is $-6.1=-61/10$, and the denominator is $(-0.1)(1.9)=(-1/10)(19/10)=-19/100$. The quotient is $610/19>0$. At $x=-1.9$, the numerator is $-5.9=-59/10$, and the denominator is $(0.1)(2.1)=(1/10)(21/10)=21/100$. The quotient is $-590/21<0$. The computed signs agree with the one-sided limits.

A Common Sign Mistake

A frequent error is to treat “the denominator tends to zero” as though zero itself had a sign. It does not. Near the input, the denominator has different signs on the two sides, and those signs must be tracked before deciding the direction of divergence.

For example, consider again $\frac{x-4}{(x+2)(x+4)}$ near $-2$. A mistaken calculation might say, “The numerator approaches $-6$, so $-6/0=+\infty$ on both sides.” This is wrong: division by zero is not defined, and the notation $-6/0$ does not encode a one-sided sign.

The correction is to factor the denominator as $(x+2)(x+4)$. Near $-2$, $x+4$ is positive and $x-4$ is negative. The factor $x+2$ is negative on the left and positive on the right. Consequently, the quotient is positive on the left and negative on the right, giving $+\infty$ from the left and $-\infty$ from the right.

Interpreting the Two Limits Together

At a simple pole, the left- and right-hand limits are infinite with opposite signs. Therefore they are not the same two-sided limit: the ordinary two-sided limit does not exist. It is still useful—and necessary—to report both one-sided limits, because each describes a distinct side of the input.

The sign procedure is concise once the factorization is clear: evaluate the numerator and the nonvanishing denominator factor at $c$, then use the sign of $x-c$ on each side. It applies precisely under the stated nonzero conditions. Those conditions distinguish this situation from cases where the numerator also vanishes or where the denominator has a higher-multiplicity zero.

Check Your Understanding

For each question, identify the relevant signs before stating the limits.

For $f(x)=\frac{x+2}{x-1}$, find the left-hand and right-hand limits at $x=1$.
For $g(x)=\frac{3-x}{(x-4)(x+1)}$, find both one-sided limits at $x=4$.
In the theorem, why must $P(c)$ and $G(c)$ both be nonzero?
If $A=P(c)/G(c)<0$, which side tends to $+\infty$, and why?
Explain why a statement such as “the denominator tends to zero” is not enough to determine the sign of an infinite limit.

Simple Poles In Rational Functions

What You'll Learn

Why the Sign on Each Side Matters

What Counts as a Simple Pole?

A Sign Rule for Both One-Sided Limits

Why the Rule Works