Limits of Polynomials and Rational Functions

Limits of Polynomial Functions

A polynomial function of degree $n$ is written as

where each $a_i$ is a real number and $a_n \neq 0$ for some natural number $n$.

We already know the fundamental limit $\lim_{x \to a} x = a$. From this, we can build up to powers of $x$. For $x^2$:

A simple induction on $n$ tells us that for any natural number $n$:

Now consider a general polynomial $f(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$. Treat each term $a_0, a_1 x, a_2 x^2, \dots, a_n x^n$ as a separate function. Using the limit laws (the limit of a sum is the sum of the limits, and the limit of a constant times a function is the constant times the limit of the function):

Make sure you understand the justification for each step: the first equality uses the sum law for limits, the second uses the constant multiple law, the third uses $\lim_{x \to a} x^k = a^k$, and the last is just the definition of $f(a)$.

Limits of Rational Functions

A rational function is a function of the form $f(x) = \frac{g(x)}{h(x)}$, where $g(x)$ and $h(x)$ are polynomials and $h(x) \neq 0$ (at least near the point of interest).

Using the quotient law for limits:

This works perfectly when $h(a) \neq 0$. But what happens when $h(a) = 0$? There are two cases to consider.

Case 1: $h(a) = 0$ and $g(a) \neq 0$

In this case, the denominator approaches zero while the numerator approaches a non-zero number. The limit does not exist (it tends to $\pm \infty$, depending on the signs).

Case 2: $h(a) = 0$ and $g(a) = 0$

This is the interesting case — the $\frac{0}{0}$ indeterminate form. Since both polynomials vanish at $x = a$, both $g(x)$ and $h(x)$ must have $(x - a)$ as a factor.

Let $k$ be the highest power of $(x - a)$ that divides $g(x)$, and let $l$ be the highest power of $(x - a)$ that divides $h(x)$. Then we can write:

where $g_1(a) \neq 0$ and $h_1(a) \neq 0$.

Now:

If $k > l$: The factor $(x - a)^{k-l}$ approaches $0$, and $\frac{g_1(x)}{h_1(x)}$ approaches $\frac{g_1(a)}{h_1(a)}$, a finite non-zero number. So:

If $k < l$: The factor $(x - a)^{k-l} = \frac{1}{(x - a)^{l-k}}$ blows up (tends to $\pm \infty$), so the limit is not defined.

If $k = l$: The factor $(x - a)^{k-l} = (x - a)^0 = 1$, so:

Theorem 2: Limit of $\frac{x^n - a^n}{x - a}$

This theorem is true even if $n$ is any rational number and $a$ is positive (a remark that will be useful later).

›Proof

We use the algebraic identity for the difference of $n$th powers:

$x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + x a^{n-2} + a^{n-1})$

Therefore, for $x \neq a$:

$\frac{x^n - a^n}{x - a} = x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + x a^{n-2} + a^{n-1}$

Now take the limit as $x \to a$. The right-hand side is a polynomial in $x$ (for fixed $a$), so its limit is its value at $x = a$:

$\begin{aligned} \lim_{x \to a} \frac{x^n - a^n}{x - a} &= \lim_{x \to a} \left( x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + x a^{n-2} + a^{n-1} \right) \\ &= a^{n-1} + a^{n-2}a + a^{n-3}a^2 + \dots + a \cdot a^{n-2} + a^{n-1} \\ &= a^{n-1} + a^{n-1} + a^{n-1} + \dots + a^{n-1} + a^{n-1} \end{aligned}$

There are exactly $n$ terms in this sum (one for each power from $0$ to $n-1$), each equal to $a^{n-1}$. Hence:

$\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$

Example 3: Applying Theorem 2

(i) $\displaystyle \lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$

We can rewrite this as:

Therefore: …