Limits of Rational Expressions

Rational Expressions Combine Several Limit Laws

A rational expression is a quotient of polynomial expressions. Its limit can often be found by first determining the limits of the polynomial parts and then applying the Limit of a Quotient Theorem. The important condition is that the denominator's limit is nonzero. When the variable is the sequence index itself, the polynomial parts need not converge, so we will also compare their degrees to analyze ratios such as $(3n^2-2n+5)/(6n^2+n+4)$.

We first establish how polynomials behave along a convergent sequence. This supplies a general substitution rule. We will then use it, along with the Limit of a Quotient Theorem, to treat rational expressions. Finally, we will study ratios of polynomials in $n$, where comparing degrees gives a useful classification.

Polynomial Expressions Along a Convergent Sequence

A real polynomial is a function of the form $P(x)=c_0+c_1x+\cdots+c_dx^d$, where $d$ is a nonnegative integer and the coefficients $c_0,\ldots,c_d$ are fixed real numbers. If $a_n\to L$, each fixed power $a_n^j$ converges to $L^j$, by the Limit of a Fixed Positive Integer Power Theorem. The Limit of a Finite Sum Theorem and the Limit of a Scalar Multiple Theorem then handle the terms of the polynomial.

Theorem (Limit of a Polynomial Expression): Let $P(x)=c_0+c_1x+\cdots+c_dx^d$ be a real polynomial. If $a_n\to L$, then $$ P(a_n)\longrightarrow P(L). $$

Proof. For each positive integer $j\leq d$, the Limit of a Fixed Positive Integer Power Theorem gives $a_n^j\to L^j$. The constant sequence $c_0$ converges to $c_0$, and the Limit of a Scalar Multiple Theorem gives $c_ja_n^j\to c_jL^j$ for every $j$. Applying the Limit of a Finite Sum Theorem to these finitely many sequences gives

P(a_n)=c_0+\sum_{j=1}^{d}c_ja_n^j \longrightarrow c_0+\sum_{j=1}^{d}c_jL^j=P(L).

When $d=0$, $P$ is the constant polynomial $c_0$, and the same conclusion follows from the Theorem (A Constant Sequence Converges). Thus the result holds for every polynomial degree. $\square$

This theorem says that a fixed polynomial can be evaluated at the limit: first take the limit of the input sequence, then evaluate the polynomial. It does not require the terms $a_n$ to be distinct, positive, or monotone.

Worked Example: A Polynomial of a Convergent Sequence

Let $a_n=2-\frac{1}{n+1}$, so $a_n\to2$, and consider $P(x)=3x^2-4x+1$. The polynomial theorem gives

3a_n^2-4a_n+1\longrightarrow 3(2)^2-4(2)+1=12-8+1=5.

The polynomial can also be checked directly. Substituting $a_n=2-\frac{1}{n+1}$ gives $3a_n^2-4a_n+1=5-\frac{8}{n+1}+\frac{3}{(n+1)^2}$, which tends to $5$. The substitution rule avoids needing this expansion each time.

Rational Expressions at a Finite Limit

A rational expression has the form $P(x)/Q(x)$, where $P$ and $Q$ are polynomials. If $a_n\to L$, the polynomial theorem gives $P(a_n)\to P(L)$ and $Q(a_n)\to Q(L)$. When $Q(L)\neq0$, the Limit of a Quotient Theorem then applies, provided the quotients are defined.

Theorem (Limit of a Rational Expression): Let $P$ and $Q$ be real polynomials, and suppose $a_n\to L$, $Q(L)\neq0$, and $Q(a_n)\neq0$ for every $n$. Then $$ \frac{P(a_n)}{Q(a_n)} \longrightarrow \frac{P(L)}{Q(L)}. $$

Proof. By the Limit of a Polynomial Expression Theorem, $P(a_n)\to P(L)$ and $Q(a_n)\to Q(L)$. The denominator limit $Q(L)$ is nonzero by assumption, and $Q(a_n)\neq0$ at every index. These are precisely the conditions needed for the Limit of a Quotient Theorem. It follows that $P(a_n)/Q(a_n)\to P(L)/Q(L)$. $\square$

The condition $Q(L)\neq0$ matters: it makes the limiting denominator suitable for the quotient law. In fact, since $Q(a_n)\to Q(L)\neq0$, the Eventual Separation from Zero Lemma also shows that $Q(a_n)$ is nonzero for all sufficiently large $n$. If an expression is defined only on a tail of the indices, the same limit argument applies to that tail.

Worked Example: Substituting a Sequence into a Rational Expression

Set $a_n=1+\frac{1}{n+2}$. Then $a_n\to1$. Consider the rational expression $\frac{a_n^2+1}{3a_n-2}$. Its denominator at the limiting input is $3(1)-2=1$, which is nonzero. Also, $3a_n-2=1+\frac{3}{n+2}>0$ for every $n\in\mathbb{N}_0$, so the expression is defined at every index. The rational-expression theorem gives

\frac{a_n^2+1}{3a_n-2} \longrightarrow \frac{1^2+1}{3(1)-2}=2.

The numerator and denominator limits are respectively $2$ and $1$; their quotient is $2$.

Ratios of Polynomials in the Index

The degree comparison for $P(n)/Q(n)$ addresses a different situation: the input $n$ does not approach a finite real number. Here $P$ and $Q$ are nonzero polynomials, and their degrees determine whether the ratio approaches zero, a nonzero constant, or has no finite limit.

To justify the comparison, it is useful to normalize a degree-$d$ polynomial by $(n+1)^d$. For each positive integer $k$, $1/(n+1)^k\to0$: given $\varepsilon>0$, choose $N$ so that $1/(N+1)<\varepsilon$; then, for $n\geq N$, $0\leq1/(n+1)^k\leq1/(n+1)<\varepsilon$. Also $n/(n+1)=1-1/(n+1)\to1$, and fixed positive integer powers preserve this limit.

Theorem (Degree Comparison for Polynomial Ratios): Let $P$ and $Q$ be nonzero real polynomials of degrees $p$ and $q$, with leading coefficients $a$ and $b$, respectively. Suppose $Q(n)\neq0$ for all sufficiently large $n\in\mathbb{N}_0$. Then:

If $p<q$, then $P(n)/Q(n)\to0$.
If $p=q$, then $P(n)/Q(n)\to a/b$.
If $p>q$, then $P(n)/Q(n)$ does not converge to a finite real number.

Proof. Define $F_n=P(n)/(n+1)^p$ and $G_n=Q(n)/(n+1)^q$. If $P(x)=a x^p+\sum_{j=0}^{p-1}a_jx^j$, then

F_n =a\left(\frac{n}{n+1}\right)^p+ \sum_{j=0}^{p-1}a_j\frac{n^j}{(n+1)^p}.

The first term tends to $a$. Each lower-degree term tends to zero because $\frac{n^j}{(n+1)^p}=(\frac{n}{n+1})^j\frac{1}{(n+1)^{p-j}}$, where the first factor tends to $1$ (and equals $1$ when $j=0$), while the second tends to zero. Hence $F_n\to a$. The same argument gives $G_n\to b$, where $b\neq0$. The Limit of a Quotient Theorem shows that $F_n/G_n\to a/b$ for all sufficiently large $n$. For those indices,

\frac{P(n)}{Q(n)} = \frac{F_n}{G_n}(n+1)^{p-q}.

If $p=q$, the final factor equals $1$, so the ratio tends to $a/b$. If $p<q$, the final factor is $1/(n+1)^{q-p}$, which tends to zero; since $F_n/G_n\to a/b$, the product tends to zero by the Limit of a Product Theorem.

If $p>q$, then $k=p-q$ is positive and $(n+1)^k\to+\infty$. Since $F_n/G_n\to a/b\neq0$, eventual sign stability and the definition of convergence imply that, for all sufficiently large $n$, $|F_n/G_n|>|a/b|/2$. Therefore

\left|\frac{P(n)}{Q(n)}\right| > \frac{|a/b|}{2}(n+1)^k

for all sufficiently large $n$. The right-hand side is unbounded, so $P(n)/Q(n)$ is unbounded on a tail. Every convergent sequence is bounded, and a sequence bounded on only finitely many initial indices and unbounded on its tail cannot be convergent. Thus there is no finite limit when $p>q$. $\square$

The hypothesis about $Q(n)$ excludes indices where the ratio is undefined. A nonzero polynomial has a nonzero leading coefficient, so the normalization argument also shows $Q(n)/(n+1)^q\to b\neq0$; consequently $Q(n)$ is indeed nonzero for all sufficiently large $n$.

Worked Example: Equal Degrees Give a Leading-Coefficient Ratio

Consider

r_n=\frac{3n^2-2n+5}{6n^2+n+4}.

Both polynomials have degree $2$, with leading coefficients $3$ and $6$. The denominator $6n^2+n+4$ is positive for every $n\in\mathbb{N}_0$, since each term is nonnegative and the constant term is positive. The degree comparison theorem therefore gives

r_n\longrightarrow\frac{3}{6}=\frac12.

The lower-degree terms affect individual values but not the limiting ratio.

Worked Example: A Lower-Degree Numerator Gives Zero

Let

s_n=\frac{5n+2}{2n^2+3}.

The numerator has degree $1$ and the denominator has degree $2$. Also, $2n^2+3>0$ for every $n$. Since the numerator's degree is smaller, the degree comparison theorem yields

s_n\longrightarrow0.

For instance, $s_0=2/3$; the limit is not a claim that every term equals zero.

Worked Example: A Higher-Degree Numerator Has No Finite Limit

Consider

t_n=\frac{2n^3-n+1}{n^2+1}.

The denominator is positive at every index, and the numerator has degree $3$, while the denominator has degree $2$. Thus the ratio has no finite limit by the degree comparison theorem. More specifically, its leading coefficients have positive ratio $2$, so the proof shows that $t_n$ is eventually positive and unbounded.

When the Denominator Limit Is Zero

The condition $Q(L)\neq0$ in the rational-expression theorem cannot be dropped. For example, let $a_n=1+\frac{1}{n+1}$, so $a_n\to1$, and consider $(a_n^2-1)/(a_n-1)$. At every index $a_n\neq1$, and factoring gives

\frac{a_n^2-1}{a_n-1}=a_n+1\longrightarrow2.

But direct substitution at the limiting input gives $(1^2-1)/(1-1)=0/0$, not a defined quotient. This does not contradict the rational-expression theorem: its denominator at the limit is zero, so its hypothesis fails. The factorization works here because $a_n-1\neq0$ at every index. In other problems, a denominator tending to zero may produce an unbounded sequence or no limit; the quotient law alone decides none of these cases.

When evaluating a rational expression, first check that the sequence is defined at the indices under consideration, then find the numerator and denominator limits separately. If the denominator limit is nonzero, apply the quotient law. If the input is the index $n$, compare polynomial degrees instead. These checks distinguish a valid substitution from a calculation that merely looks plausible.

Check Your Understanding

Use the polynomial and rational-expression theorems, including their hypotheses, to answer these questions.

If $a_n\to-2$, what is the limit of $a_n^3+2a_n-1$?
For a rational expression $P(a_n)/Q(a_n)$, why is $Q(L)\neq0$ needed when $a_n\to L$?
What is the limit of $(4n^2+3)/(8n^2-n+2)$, and which features determine it?
Does $(n+1)/(n^3+1)$ tend to zero, a nonzero constant, or fail to have a finite limit?
Why does the quotient limit law not directly evaluate $(a_n^2-1)/(a_n-1)$ when $a_n\to1$?

What You'll Learn