Regardless of your early surrounding or schooling background, we know for one that there are two kinds of *mathematical objects* that are kind of hard to miss in life. The names? **Polynomial** and **infinity**! While the former might have sounded a bit like the name of a *snake*, polynomials is a *one-of-its-kind* mathematical entity whose perfection defies our mathematical imagination.

For one, polynomials are well-known for being *infinitely smooth* and *never-ending*, while at the same time, they could be a *line*, a *parabola*, or any kind of weird, **infinitely-malleable curve** ready to assume any shape drawn without *lifting* the pencil (kind of). Heck, polynomials are a favorite object of *platonic desire* among math enthusiasts. Talk about the *interaction* between polynomial and infinity!

So with all that goodness, it makes sense for us to inquire a bit as to why despite of having similar forms, the behaviors of polynomials at the infinities differ, leading to some seemingly unrelated insights about their **properties** in general.

All right. Enough said. Time to buckle the seat belt, and let the theoretical musing begins!

## Polynomials — A Review

A *non-zero*, real-valued function of the form $\displaystyle cx^n$ — where $c \in \mathbb{R} \setminus \{ 0 \}$ and $n \in \mathbb{N}$ — forms the building blocks of polynomials. For that reason, they are usually called the (non-zero)** monomials**, with the number $n$ being the **degree** of the monomial in question. For example, the *constant function* $\frac{\pi}{4}$ represents a monomial with degree $0$, whereas the function $ex^{666}$ would be a monomial with degree $666$ (!).

(for the record, the *zero function* also falls into the category of monomials, except that its degree is usually left *undefined*, for the obvious reason that it can be expressed *numerous* forms).

Once there, we can define a **non-zero polynomial** as a function of the form:

\begin{align*} f_1(x)+ \dots + f_m(x) \end{align*}

where $m \in \mathbb{N_+}$ (i.e., a *positive* integer) and each $f_i(x)$ is a *non-zero* monomial.

In which case, we define the **degree** of a *non-zero* polynomial as the *maximal* degree of its monomials (always well-defined, since there’s only a *finite* number of monomials). For example, the constant function $\frac{\ln 2 + 3}{2e}$ represents a polynomial of degree $0$, and the function $3 + 2x^3 + x + x^2$ a polynomial of degree $3$.

(as with before, the *zero function* still falls into the category of polynomials, with the caveat that its degree is usually left undefined — due to the ambiguity of its leading term)

For the sake of *simplicity* and *consistency* though, we usually prefer to write a *non-zero* polynomial by gathering up the *like terms*, and rearranging the resulting terms so that the monomials are presented in increasing/decreasing order based on their degrees. Once that’s done, we refer to the monomial of the highest degree as the **leading term** of the polynomial, and its coefficient the **leading coefficient** of the polynomial.

For example, instead of writing $10x^3+2x^2+ 5x+ 6x^2 + 5 + 2 x^3$ as our polynomial, we prefer to regroup the like terms and rewrite it as $12x^3 + 8x^2 + 5x + 5$, from which it can be seen that we have a polynomial of *degree* $3$ (i.e., a **cubic polynomial**) — with $12$ as the *leading coefficient*.

In general, a *non-zero* polynomial can have *any* non-zero number as leading coefficient. However, in the special case where the leading coefficient is $1$, it is as if it it kind of disappears out of sight. In which case, we refer to the said polynomial as a **monic polynomial**.

In addition, since every *non-zero* polynomial can be re-expressed in terms of* monic polynomial* by *factoring out* the leading coefficient, it sometimes makes sense to analyze the properties of a non-zero polynomial by looking instead at the properties of its *underlying monic polynomial*.

Alternatively, we can also define a polynomial *recursively* as follows:

Definition 1 — Recursive Definition of Polynomial

*A real-valued* function $p(x)$ is a **polynomial** if and only if it satisfies one of the two following options — through a *finite* number of iterations:

- $p(x)=c$ for some $c \in \mathbb{R}$ (i.e., a
*constant function*). - $p(x) = x q(x) + c$, where $q(x)$ is a
*polynomial*and $c \in \mathbb{R}$.

For example, $x^2 + 2x + 1$ is a polynomial by the standard of this *recursive definition*, since $x^2+2x+1 = x ( x+2) + 1$, where $x+2$ is very well a polynomial (why?) and $1$ very well a constant. In general, the recursive definition of polynomial suggests a way of *reducing* a polynomial of degree $n+1$ to that of degree $n$ — an handy insight when it comes to proving facts about *non-zero* polynomials using the so-called **mathematical induction** on the *degree* of polynomials.

## A Venture Into the Infinite Limits

When tackling the behavior of a function at the infinities, it makes sense that we equip ourselves with some **limit laws** on how *infinity-converging* and *constant-converging* functions interact with each other. To that end, we present the following *6* sets of limit laws for your pleasure. 🙂

### Interactions Between Infinity-Converging and Constant Functions

To start off, let’s imagine that we have a function $f(x)$ and a constant function $c$, where $f \to \infty$ as $x \to \infty$. The question is, what does the function $f(x) + c$ tends to — if any — as $x \to \infty$?

Here, it shouldn’t be a surprise that $f(x)+c$ converges to $\infty$ as well. To see why, we first note that since $f \to \infty$ as $x \to \infty$, the **definition of limit** implies that:

Given

anynumber — howeverlarge— there’ll always besomeneighborhood$N$ of the form $(\bigcirc, \infty)$, such that every member in $f(N)$exceedsthis number.

This means that given *any* number $r$, it would then be possible to find one such *neighborhood* $N$ such that $f(x) > r – c$ for all $x \in N$. In which case, we would have that:

\begin{align*} f(x) + c > r \qquad (\forall x \in N) \end{align*}

thereby showing that the function $f(x) + c$ increases *beyond bound* as $x \to \infty$ — regardless of the value of $c$.

Naturally, this leads to the following question: what if $f$ actually converges to $-\infty$ instead as $x \to \infty$? Well, referring to the *definition of limit* again, we have that:

Given

anynumber — howevernegative— there’ll always besomeneighborhood$N$ of the form $(\bigcirc, \infty)$, such that every member in $f(N)$precedesthis number.

This means that given *any* number $r$, it would then be possible to find one such *neighborhood* $N$ such that $f(x) < r – c$ for all $x \in N$. In which case, we would have that:

\begin{align*} f(x) + c < r \qquad (\forall x \in N) \end{align*}

hence showing that the function $f(x) + c$ decreases *beyond bound* $x \to \infty$, and this again regardless of the *sign* and *size* of $c$. Putting everything together, we obtain our *first* set of limit laws concerning **constant functions**:

Proposition 1 — Limit Laws Concerning Constant Functions (Sum)

Given a *real-valued* function $f(x)$ and a constant function $c$, if as $x \to \Box$ (where $\Box$ could be a number, $+\infty$ or $-\infty$), $f(x)$ converges to one of the infinities, then the function $f(x)+c$ converges to the *same* infinity $f(x)$ converges to — as $x \to \Box$.

Now, what about the cases where a function is *multiplied* by a constant, as in the case of $c f(x)$ where $c>0$? Well, as logic would have dictated, the **convergence behavior** of $c f(x)$ should depend on what $f(x)$ converges to in the first place.

For example, if we know in advance that $f(x)$ converges to $\infty$ as $x \to \infty$, then the definition of limit would imply that given *any* number $r$ — no matter how *positive* — we can always find a neighborhood $N$ of the form $(\bigcirc,\infty)$ such that:

\begin{align*} f(x) & > \frac{r}{c} \qquad (\forall x \in N) \end{align*}

That is,

\begin{align*} cf(x) & > r \qquad (\forall x \in N) \end{align*}

showing that the function $cf(x)$ increases *without bound* as $x \to \infty$.

On the other hand, if $f(x) \to -\infty$ instead as $x \to \infty$ (remember that $c>0$), then we can similarly infer that given *any* number $r$ — however *negative* — it is always possible to find a neighborhood $N$ of the form $(\bigcirc,\infty)$ such that:

\begin{align*} f(x) & < \frac{r}{c} \qquad (\forall x \in N) \end{align*}

That is,

\begin{align*} cf(x) & < r \qquad (\forall x \in N) \end{align*}

showing that the function $cf(x)$ decreases *without bound* as $x \to \infty$.

Now, this still leaves us with the scenario where $c$ is *negative*. In which case, using very similar arguments, it can still be shown that as $x \to \infty$:

- $f(x) \to +\infty$ implies that $cf(x) \to -\infty$.
- $f(x) \to -\infty$ implies that $cf(x) \to +\infty$.

So that if we combine these insights together, we get our *second* set of limit laws concerning **constant functions**:

Proposition 2 — Limit Laws Concerning Constant Functions (Product)

Given a *real-valued* function $f(x)$ and a constant function $c$, if as $x \to \Box$ (where $\Box$ could be a number, $+\infty$ or $-\infty$), $f(x)$ converges to one of the infinities, then the following cases apply as $x \to \Box$:

- If $c>0$, then $c f(x)$ converges to the
*same*infinity $f(x)$ converges to. - If $c<0$, then $c f(x)$ converges to the
*opposite*infinity $f(x)$ converges to.

Interestingly, the above limit laws applies *irrespective of the size of $c$*. For example, $c$ could be $0.00001$ and $f(x)$ could converge to $+\infty$, and $cf(x)$ would have converged to $+\infty$ regardless. Likewise, $c$ could be $\frac{1}{e}$ and $f(x)$ could converge to $-\infty$, and $cf(x)$ would have converged to $-\infty$ regardless.

### Infinity + Constant

Moving from the constant functions into the general realm, let’s assume that we’re given two functions $f(x)$ and $g(x)$, such that as $x \to \infty$, $f$ converges to one of the infinities and $g$ converges to some real number $G$. The question is, what can we extrapolate about the **convergence behavior** of $f+g$ — if any?

Here, to figure this out, we begin by noticing that since $g$ converges to an *actual* number $G$, by virtue of the *definition of limit*, there must be a neighborhood $N$ of the form $(\bigcirc, \infty)$ such that:

\begin{align*} G- 1 < g(x) < G+1 \qquad (\forall x \in N)\end{align*}

In other words, the function $f + g$ has the *peculiar* property that:

\begin{align*} f(x) + (G- 1) < f(x) + g(x) < f(x) + (G+1) \qquad (\forall x \in N)\end{align*}

So that if $f(x) \to \infty$ as $x \to \infty$, then so does the function $f(x) + (G-1)$ (why?). In which case, an application of the **Squeeze Theorem** would show that *the same is true* for the function $f +g$ as well.

In a similar-but-opposite manner, if $f(x) \to -\infty$ as $x \to \infty$ instead, then the same applies to the function $f(x) + (G+1)$. In which case, invoking the *Squeeze Theorem* again would show that *the same is true* for the function $f + g$ as well, so that if we combine these facts together, we now have a *third* set of limit laws — this time concerning the **sum of infinity-converging and constant-converging functions**:

Proposition 3 — Limit Laws Concerning Infinity-Converging and Constant-Converging Functions (Sum)

Given two real-valued functions $f(x)$ and $g(x)$, if as $x \to \Box$ (where $\Box$ could be a number, $-\infty$ or $-\infty$), we have both that:

- $f$ tends to one of the infinities.
- $g$ tends to a real number.

then the function $f + g$ tends to the *same* infinity $f$ tends to — as $x \to \Box$.

Or more concisely,

\begin{align*} \infty + c & = \infty & -\infty + c & = -\infty \end{align*}

where the first identity holds even when $c$ is *extremely negative*, and the second even when $c$ is *extremely positive.*

### Infinity × Constant

Now, if the two functions $f(x)$ and $g(x)$ are such that as $x \to \infty$, $f(x)$ converges to one of the infinities and $g$ converges to some *real *number $G$, what can we say about the **convergence behavior** of the function $fg$ — if any?

Here, it would only seem logical that the convergence behavior of $fg$ depend on the **sign** of $G$. For example, if we know in advance that $G$ is *positive*, then the *definition of limit* would dictate that there be a neighborhood $N_1$ of the form $(\bigcirc, \infty)$ such that:

\begin{align*} \frac{G}{2} = G – \frac{G}{2} < g(x) < G + \frac{G}{2} = \frac{3G}{2} \qquad (\forall x \in N_1) \end{align*}

so that if $f(x) \to \infty$ as $x \to \infty$, then there must be a neighborhood $N_2$ of a similar form where $f(x)>0$. In which case, *multiplying the two inequalities* would show that the function $fg$ satisfies the following *peculiar* property:

\begin{align*} f(x) \ \frac{G}{2} < f(x) g(x) < f(x) \ \frac{3G}{2} \qquad (\forall x \in N_1 \cap N_2) \end{align*}

Once there, it can be readily seen by **Squeeze Theorem** that as $x \to \infty$, the function $fg$ converges to the *same* infinity $f(x)$ converges to. Now, that was *if* we know that $f(x) \to +\infty$, but if $f(x) \to -\infty$ instead as $x \to \infty$, then an *analogous* line of reasoning would take us to the same conclusion anyway.

On the other hand, if $G$ were *negative* instead, then by our previous remark, as $x \to \infty$, the function $f[-g]$ would converge to the *same* infinity $f$ converges to, which means that *by extension*, as $x \to \infty$, the function $fg$ — which is equivalent to the function $- f [-g]$ — would converge to the *opposite* infinity $f(x)$ converges to. That takes care of the case where $G<0$.

Now, as for the *rare* case where $G=0$, it turns out that there is really not much we can say about the limit of $fg$. In fact, it can be readily seen that as $x \to \infty$:

- $x \to \infty$ and $\displaystyle \frac{1}{x} \to 0$, but the function $\displaystyle x \cdot \frac{1}{x} \to 1$.
- $x \to \infty$ and $\displaystyle \frac{1}{x^2} \to 0$, but the function $\displaystyle x \cdot \frac{1}{x^2} \to 0$.
- $x^2 \to \infty$ and $\displaystyle \frac{1}{x} \to 0$, but the function $\displaystyle x^2 \cdot \frac{1}{x} \to \infty$.

In other words, the function $fg$ can literally converge to *anything* when $G=0$! And with that last case settled, we can now synthesize all the findings into our *fourth* set of limit laws — this time concerning the **product of infinity-converging and constant-converging functions**:

Proposition 4 — Limit Laws Concerning the Infinity-Converging and Constant-Converging Functions (Product)

Given two real-valued functions $f(x)$ and $g(x)$, if as $x \to \Box$ (where $\Box$ could be a number, $+\infty$ or $-\infty$), $f$ converges to one of the infinities and $g$ converges to a real number $G$, then the following cases apply as $x \to \Box$:

- If $G>0$, then the function $fg$ converges to the
*same*infinity $f$ converges to. - If $G<0$, then the function $fg$ converges to the
*opposite*infinity $f$ converges to. - If $G=0$, then no information can be extracted about the limit of $fg$.

In other words,

\begin{align*} \pm \infty \cdot + & = \pm \infty & \pm \infty \cdot – & = \mp \infty \end{align*}

where both identities hold regardless of the *magnitude* of $G$.

### Infinity + Infinity

When the functions $f(x)$ and $g(x)$ both tends to one of the infinities, interesting *interactions* are bound to happen. For example, if we know that *both* $f$ and $g$ converge to $\infty$ as $x \to \infty$, then the *definition of limit* would guarantee that there be a neighborhood $N$ of the form $(\bigcirc, \infty)$ such that:

\begin{align*} f(x)+g(x) > f(x) + 0 \qquad (x \in N) \end{align*}

from which we can infer by **Squeeze Theorem** that the function $f + g$ converges to $\infty$ as well — as $x \to \infty$.

Similarly, if *both* $f$ and $g$ converge to $-\infty$ as $x \to \infty$, then the *definition of limit* would again guarantee the existence of a neighborhood $N$ of the form $(\bigcirc, \infty)$ such that:

\begin{align*} f(x)+g(x) < f(x) + 0 \qquad (x \in N) \end{align*}

in which case, we can infer — again by *Squeeze Theorem* — that the function $f + g$ converges to $-\infty$ as well — as $x \to \infty$.

On the other hand, if $f$ and $g$ converge to *opposite infinities* instead, then it would transpire that there is really not much we can say about the limit of $f + g$, as it can be readily seen that as $x \to \infty$:

- $\displaystyle x + 1 \to +\infty$ and $\displaystyle -x \to -\infty$, but $\displaystyle (x + 1) + (-x) \to 1$.
- $\displaystyle 2x \to +\infty$ and $\displaystyle -x \to -\infty$, but $\displaystyle (2x) + (-x) \to \infty$.
- $\displaystyle x \to +\infty$ and $\displaystyle -2x \to -\infty$, but $\displaystyle x + (-2x) \to -\infty$.

In any case, putting all the findings together, we see that the *fifth* set of limit laws — this time concerning the **sum of infinity-converging functions** — is now in order:

Proposition 5 — Limit Laws Concerning Infinity-Converging Functions (Sum)

Given two real-valued functions $f(x)$ and $g(x)$, if as $x \to \Box$ (where $\Box$ could be a number, $+\infty$ or $-\infty$), both $f$ and $g$ converge to the *same* kind of infinity, then the same applies to the function $f + g$.

However, if $f$ and $g$ converge to *opposite* infinities, then no information can be extracted about the limit of $f+g$.

Or more concisely,

\begin{align*} \infty + \infty & = \infty & -\infty + -\infty & = -\infty \end{align*}

Enjoying those *squiggly symbols* so far? 🙂

### Infinity × Infinity

Now, suppose that the functions $f(x)$ and $g(x)$ both converge to one of the infinities, then is it *always* possible to extrapolate something about the limit of the function $fg$? As it turns out, the answer is a resounding *yes*, with the caveat being that this limit varies a bit — depending on whether $f$ and $g$ converge to the *same* infinity or *opposite* infinities.

For example. if both $f$ and $g$ converge to $\infty$ as $x \to \infty$, then the *definition of limit* would guarantee that there be a neighborhood $N$ of the form $(\bigcirc, \infty)$ such that:

\begin{align*} f(x)g(x) & > f(x) \cdot 1 \qquad (\forall x \in N)\end{align*}

from which we can infer by **Squeeze Theorem** that the function $fg$ *increases* beyond bound as $x \to \infty$.

And if both $f$ and $g$ converge to $-\infty$ as $x \to \infty$, then our previous finding would again show that $fg$, which is equivalent to $(-f)(-g)$, converges to $\infty$ all the same — as $x \to \infty$.

As for the cases where $f$ and $g$ converge to *opposite infinities*, recycling the **“negating-the-function” trick** would yield that the function $fg$ now converges to $-\infty$ instead. For example, if as $x \to \infty$, $f \to +\infty$ and $g \to -\infty$, then we can infer that $f(-g) \to +\infty$, which means that the function $fg$ converges to $-\infty$ as $x \to \infty$.

And with all these cases settled, we can now move on into synthesizing the findings into our *sixth* and last set of limit laws — this time concerning the **product of infinity-converging functions**:

Proposition 6 — Limit Laws Concerning Infinity-Converging Functions (Product)

Given two real-valued functions $f(x)$ and $g(x)$, if as $x \to \Box$ (where $\Box$ could be a number, $+\infty$ or $-\infty$), both $f$ and $g$ converge to one of the infinities, then the following cases apply as $x \to \Box$:

- If both $f$ and $g$ converge to the
*same*infinity, then $fg \to +\infty$. - If $f$ and $g$ converge to
*opposite*infinities, then $fg \to -\infty$.

Or more concisely,

\begin{align*} \infty \cdot \infty & = \infty & \infty \cdot -\infty & = -\infty \\ -\infty \cdot -\infty & = \infty & -\infty \cdot \infty & = -\infty \end{align*}

## Behaviors of Polynomials at the Infinities

All right. Now that we have all those limit laws in our *bag of tricks*, we can proceed with confidence into tackling the **end-behaviors** of all polynomials: first with the *monomials*, and then the polynomials in general.

### Monomials

For the **constant functions** (*zero function* included), the end-behaviors are trivial. As for the monomials with degree $1$ or more, we begin by noticing that:

- The function $x$ goes from $-\infty$ to $+\infty$.
- The function $x^2$ goes from $+\infty$ to $+\infty$.
- The function $x^3$ goes from $-\infty$ to $+\infty$.
- $\ldots$

In fact, by using **mathematical induction** (coupled with some *limit laws on infinities*), we can show that for all *monic* monomials with degree $1$ or more:

- If the monomial has an
*odd*degree, then it goes from $-\infty$ to $+\infty$. - If the monomial has an
*even*degree, then it goes from $+\infty$ to $+\infty$.

Generalizing a bit further, we see that the end-behaviors of a *non-constant monomial* (i.e., $cx^n$, where $c \ne 0$ and $n \ge 1$) — in general — depends on both the **parity** of the monomial and the **sign** of the coefficient $c$:

**Odd-degree Monomial**: If $c>0$, then by the*limit law on constant functions*, $cx^n$ must share the same end-behaviors as $x^n$ — going from $-\infty$ to $+\infty$. On the other hand, if $c<0$, then $cx^n$ goes from $+\infty$ to $-\infty$.**Even-degree Monomial**: If $c>0$, then the*limit law on constant functions*again implies that $cx^n$ must share the same end-behaviors as $x^n$ — now going from $+\infty$ to $+\infty$. On the other hand, if $c<0$, then $cx^n$ goes from $-\infty$ to $-\infty$.

And with that, our analysis on the **end-behaviors of monomials** is now complete:

Theorem 1 — End-Behaviors of Monomials

The behaviors of a monomial $m(x)$ at the infinities can be analyzed based on its *degree* and the *sign* of its coefficient:

- If $m(x)$ is a
**constant**, then its end-behaviors are trivial. - If $m(x)$ is
**odd**, then:- If the coefficient is
*positive*, then $m(x)$ goes from $-\infty$ to $+\infty$. - If the coefficient is
*negative*, then $m(x)$ goes from $+\infty$ to $-\infty$.

- If the coefficient is
- If $m(x)$ is
**even**(excluding the*constant functions*), then:- If the coefficient is
*positive*, then $m(x)$ goes from $+\infty$ to $+\infty$. - If the coefficient is
*negative*, then $m(x)$ goes from $-\infty$ to $-\infty$.

- If the coefficient is

which — considering that monomials come in all different shapes and forms — is a pretty amazing achievement. 🙂

### General Polynomials

With the end-behaviors of monomials settled, it’s only a few steps before we figure out the general end-behaviors of *all* polynomials. Again, similar to the case with monomials, if the polynomial is a *constant*, then its end-behaviors are trivial. As for the other polynomials of the form $\displaystyle c_n x^n + \dots + c_0 x^0$ ($n \ge 1, c_n \ne 0$), we can factor out the **leading term**, thereby producing to the following equality:

\begin{align*} c_n x^n + \dots + c_0 x^0 = c_n x^n \left( 1 + \dots + \frac{c_0 x^0}{c_n x^n} \right) \qquad (x \ne 0) \end{align*}

Once there, leveraging the fact that $\displaystyle 1 + \dots + \frac{c_0 x^0}{c_n x^n}$ tends to $1$ as $x$ tends to $+\infty$ or $-\infty$, we can see that $c_n x^n + \dots + c_0 x^0$ essentially has the same convergence behavior as $c_n x^n$ — at the* infinities*. That is, the contribution of the **lower terms** becomes *invariably insignificant* as we get near the infinities — an impressive finding which we shall incorporate into our *theorem of the day*!

Theorem 2 — End-Behaviors of Polynomials

The behaviors of a polynomial at the infinities can be broken down into the following two cases:

- For a
**constant polynomial**, the end-behaviors are trivial. - Otherwise, it shares the
*same*end-behaviors as that of its**leading term**.

Essentially, we get to know about the end-behaviors of *any* polynomial we can come up with, without needing to know how to solve for the **roots**, or how the polynomial behaves *within* the infinities! For example, we could be given the function $(x-3)(x-2)(x-1)x(x+1)(x+2)(x+3)$, which has a bit of things going on in between the infinities:

However, knowing that we have a polynomial of degree $7$ with *positive* coefficient, we can determine — with almost zero computation — that the function essentially travels from $-\infty$ to $+\infty$, even without having its graph at our disposal! Now, how is that in for a treat? 🙂

### Ramifications

As can be seen with our earlier finding, an **odd-degree polynomial** always goes from one kind of infinity to the other, whereas an **even-degree polynomial** (save the *constant polynomials*) stays with the same kind of infinity. This leaves us with *two* kinds of polynomials with fundamentally-different end-behaviors.

In particular, given an *odd-degree* polynomial $p(x)$, the fact that it goes from one infinity to the other suggests that there must be *two* *neighborhoods* — one of the form $(-\infty, a]$, and the other of the form $[b, +\infty)$ — such $p(x)>0$ in one, and $p(x)<0$ in the other. As a result, we can see that the number $0$ — which is an *intermediate value* — must have been attained somewhere in the interval $[a,b]$ according to the **Intermediate Value Theorem**. Put it simply, an *odd-degree* polynomial always has *at least* one real root — regardless of the complexity of its shape and form.

In fact, more is true. Using a very similar argument, it can be shown that for every *odd-degree polynomial* has the peculiar property that *every* number — however large or small — constitutes an *intermediate value*, and hence must be attained somewhere in the polynomial’s graph according to the *Intermediate Value Theorem*. Algebraically, this means that the equation $p(x)=c$ always has *at least* one solution — even if such solutions can be hard to determine or approximate through *algebraic/**numerical* techniques of **root finding**.

As an illustration, the **quintic polynomial** $-\frac{\pi}{3} x^5 – 2x^4 + 123 x^3 – x + 5$, which we now know goes from $+\infty$ to $-\infty$, must have a *root* somewhere in its graph — even if we might have very little clue as to how to find it. In fact, what we have here is a polynomial that actually attains *every* single real number in its graph, effectively mapping the set of real numbers *to the set of real numbers*!

Moving on to the other end of the spectrum, an *even-degree* polynomial (excluding the *constant polynomials*), which takes the shape of either a **cap** or a **cup**, must either attain a **maximum** in its graph, or a **minimum** in its graph. However, what’s more interesting is that *if* such a polynomial attains a *maximum*, then any number *below* the maximum represents an *intermediate value*, and hence must be attained somewhere in the graph of the polynomial. Similarly, if the polynomial in question attains a *minimum* instead, then any value *beyond* the minimum constitutes an *intermediate value*, and thus must be attained at some point in the graph of the polynomial as well.

For example, the **quartic polynomial** $x^4 – x -\frac{1}{2}$, which we know goes from $+\infty$ to $+\infty$, must have — by extension of this reasoning — attained a *minimum* $m$ somewhere in its graph, for *if that’s not the case*, then it would be possible to construct a sequence $x_i$ — which converges to either a real number, $+\infty$, or $-\infty$ — such that the sequence $f(x_i)$ converges to $-\infty$, thereby contradicting the *continuity* and *end-behaviors* of the polynomials in question.

Now, we do want to emphasize that our previous “proof” on the existence of the minimum $m$ is neither *trivial* nor *constructive*, in the sense even if we acknowledge the existence of the minimum, it’s not always possible to find the *exact* location where $m$ is attained through *algebraic* or *numerical* methods. But whatever this $m$ is, what we do know is that *each* value beyond $m$ is also attained somewhere in the graph, so that even if we might be clueless as to *where* exactly those values are attained, we are nevertheless correct in asserting that the quartic polynomial $x^4 – x -\frac{1}{2}$ maps the set of real number to the interval $[m,\infty)$.

Bottom line? It seems that regardless of whether the polynomial is *even* or *odd*, there’s always something we can comment about!

## Afterwords

Wow! That was a fruitful venture with plenty of ideas and insights about **infinite limits** and **polynomials**! While it definitely leaves a lot more to be said about polynomials and their other exotic properties, the fact that we’ve mapped out their **end-behaviors** and in the process, went through a whole bunch of **limit laws** involving the infinities, suggests that we have already covered a lot of ground — which hopefully represents some intellectual progress as well. 🙂

So without further ado, here’s an *interactive table* summarizing all our findings so far:

Monomial |

Degree (Monomial) |

Constant Function |

Zero Function |

Non-Zero Polynomial |

Degree (Non-Zero Polynomial) |

Leading Terms |

Leading Coefficient |

Monic Polynomial |

Polynomial (Recursive Definition) |

Mathematical Induction |

Given two *real-valued* functions $f(x)$, $g(x)$ and a constant function $c$, if we take the limit as $x$ tends to $\Box$ (where $\Box$ can be either a number, $+\infty$ or $-\infty$), then the following 6 sets of **limit laws** apply**:**

If $f(x) \to +\infty$, then so does the function $f(x)+c$. Similarly, if $f(x) \to -\infty$, then so does the function $f(x)+c$.

If $c>0$ and $f(x)$ converges to one of the infinities, then $c f(x)$ converges to the *same* infinity $f(x)$ converges to.

If $c<0$ and $f(x)$ converges to one of the infinities, then $c f(x)$ converges to the *opposite* infinity $f(x)$ converges to.

If $f$ tends to one of the infinities and $g$ tends to a real number, then $f+g$ will tend to the infinity $f$ tends to.

If $f$ tends one of the infinities, and $g$ tends to a real number $G$, then the following cases apply:

- If $G>0$, then $fg$ converges to the
*same*infinity $f$ converges to. - If $G<0$, then $fg$ converges to the
*opposite*infinity $f$ converges to. - If $G=0$, then no information can be extracted about the limit of $fg$.

If both $f$ and $g$ converge to the *same* kind of infinity, then so does the function $f + g$.

If $f$ and $g$ converge to *opposite* infinities, then no information can be extracted about the limit of $f+g$.

If both $f$ and $g$ converge to the *same* infinity, then $fg \to +\infty$.

If $f$ and $g$ converge to opposite infinities, then $fg \to -\infty$.

The behaviors of a monomial $m(x)$ at the infinities can be analyzed based on its *degree* and the *sign* of its coefficient:

- If $m(x)$ is a
**constant**, then its end-behaviors are trivial - If $m(x)$ is
**odd**, then:- If the coefficient is
*positive*, then $m(x)$ goes from $-\infty$ to $+\infty$. - If the coefficient is
*negative*, then $m(x)$ goes from $+\infty$ to $-\infty$.

- If the coefficient is
- If $m(x)$ is
**even**(excluding the*constant functions*), then:- If the coefficient is
*positive*, then $m(x)$ goes from $+\infty$ to $+\infty$. - If the coefficient is
*negative*, then $m(x)$ goes from $-\infty$ to $-\infty$.

- If the coefficient is

The behaviors of a polynomial at the infinities can be broken down into the following two cases:

- For a
**constant polynomial**, the end-behaviors are trivial. - Otherwise, it shares the
*same*end-behaviors as that of its**leading term**.

A polynomial of **odd degree** — which goes from one infinity to the other — must attain *every* single real number in its graph. As a result, it maps the set of real numbers to itself, and always have at least a root — even if such roots could be hard to determine or approximate using *algebraic*/*numerical* methods.

A polynomial of **even degree** — which stays with the same infinity — either attains a maximum $M$ (if it takes the shape of a **cap**), or a minimum $m$ (if it takes the shape of a **cup**). In the first scenario, the polynomial maps the set of real number to $(-\infty, M]$, and in the second scenario, to $[m, \infty)$.

So there you have it! A little venture into polynomials and infinities on the juncture of calculus and real analysis. By the way, do you have a **favorite polynomial** of your own? If so, be sure to take a good new look at it using what you’ve just known!