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!

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.

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. 🙂

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 any number — however large — there’ll always be some neighborhood $N$ of the form $(\bigcirc, \infty)$, such that every member in $f(N)$ exceeds this 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 any number — however negative — there’ll always be some neighborhood $N$ of the form $(\bigcirc, \infty)$, such that every member in $f(N)$ precedes this 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.

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.

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 $G$.

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? 🙂

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*}

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.

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

e 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 $-\infty$.

: 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.

, 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 $-\infty$.

, then: 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 $-\infty$.

(excluding the constant functions), then:

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

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

e 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

e 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? 🙂

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.

An odd-degree polynomial is like God — passing through every single height without leaving a trace! Click to Tweet

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!

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:

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!

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