How can I prove that a polynomial with degree $n$ is continuous everywhere in $\mathbb{R}$ using definitions?

With induction. I can show that this polynomial is continuous at $x_0$ but I do not know how to prove that it is continuous everywhere in $\mathbb{R}$. I know that a function is continuous on an interval if it is continuous at every point on the interval but this implies that I need to show for every $x_0\in\mathbb{R}$, the polynomial is continuous.

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    $\begingroup$ Related: math.stackexchange.com/questions/314552/… $\endgroup$ Commented Mar 30, 2014 at 4:55
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    $\begingroup$ Your question is not clear. Saying "I can show that this polynomial is continuous at $x_0$", which $x_0$ do you mean, it is not mentioned in the statement you want to prove. If it is an arbitrary real number, your proof is complete. Did you maybe mean instead that you proved continuity at $x=0$? $\endgroup$ Commented Mar 30, 2014 at 5:38
  • $\begingroup$ No, I meant arbitrary $x_0$. $\endgroup$
    – user48876
    Commented Mar 30, 2014 at 7:34

3 Answers 3

  1. $f(x)=x$ is continuous everywhere
  2. If $f(x)$ and $g(x)$ are continuous in $D$ then $f(x)g(x)$ in continuous on $D$.
  3. Using 2 and 1 show that $x^n$ is continuous for every $n\in \mathbb{N}$
  4. If $f(x)$ and $g(x)$ are continuous on $D$ then $f(x)+g(x)$ is continous on $D$
  5. Now use 3 and 4.
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    $\begingroup$ Hi, I've follow your steps and would like to check if my proof is okay 1)elementary proof 2)algebraic property of coninuous functions proof 3) Proof: If f1(x)=x and f2(x)=x are both continuous on D then f1(x)f2(x)=x^2 is continuous. Hence we can conclude that f1(x)....fm(x)=x^n, for n=m(m+1)/2 is continuous on D. 4)algebraic property of continuous functions 5)proof: from 3) we know that x^n is contionuous on D and hence if we add a polynomial of degree n-1 or smaller then by 4) we can conclude that x^n+x^(n-1)+...+1 is continuous. Hence all polynomials are continuous. $\endgroup$
    – user634512
    Commented Mar 29, 2019 at 18:15
  • $\begingroup$ @ThePoorJew. Yes looks okay $\endgroup$
    – hrkrshnn
    Commented Apr 1, 2019 at 13:21

Using the definition of a limit it is easy to prove (and you should prove as a simple exercise) that $$\lim_{x \to a}k = k, \lim_{x \to a}x = a\tag{1}$$ Using these results and induction and laws of algebra of limits it is an easy matter to prove that any polynomial function $f$ with real coefficients is continuous everywhere.

We will use induction on the degree of a polynomial. Polynomials of degree $0$ are constants and by first result in $(1)$ a constant is continuous everywhere. Now we assume that any polynomial of degree less than $n$ is continuous everywhere and let $f$ be a polynomial of degree $n$. Let $a$ be any arbitrary real number. Let $$f(x) = a_{0}x^{n} + a_{1}x^{n - 1} + \cdots + a_{n - 1}x + a_{n}$$ and then we have \begin{align} \lim_{x \to a}f(x) &= \lim_{x \to a}a_{0}x^{n} + a_{1}x^{n - 1} + \cdots + a_{n - 1}x + a_{n}\notag\\ &= \lim_{x \to a}x(a_{0}x^{n - 1} + a_{1}x^{n - 2} + \cdots + a_{n - 1}) + a_{n}\notag\\ &= \lim_{x \to a}xg(x) + a_{n}\notag\\ &\,\,\,\,\,\,\,\,\text{(note that }g(x)\text{ is a polynomial of degree }(n - 1))\notag\\ &= \lim_{x \to a}x\cdot\lim_{x \to a}g(x) + \lim_{x \to a}a_{n}\text{ (laws of algebra of limits)}\notag\\ &= ag(a) + a_{n}\text{ (}g(x)\text{ is continuous and using (1))}\notag\\ &= a(a_{0}a^{n - 1} + a_{1}a^{n - 2} + \cdots + a_{n - 1}) + a_{n}\notag\\ &= a_{0}a^{n} + a_{1}a^{n - 1} + \cdots + a_{n - 1}a + a_{n}\notag\\ &= f(a)\notag \end{align} Hence polynomial $f(x)$ is continuous at $a$. Since $a$ was an arbitrary real number it follows that $f(x)$ is continuous everywhere. The proof is now complete by principle of mathematical induction and every polynomial with real coefficients is continuous everywhere.


Lemma 1: Limit of power function is the corresponding value of the function itself. $$ \begin{align} \lim_{x\to a}a_ix^i&=a_i\lim_{x\to a}x^i\\ &=a_i\prod_{j=1}^{i}\lim_{x\to a}x\tag{product rule of limit}\\ &=a_ia^i \end{align} $$

  1. Every polynomial function is continuous Let the polynomial be $P(x)=a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n=\sum_{i=0}^n a_ix^i$ $$ \begin{align} \lim_{x\to a}P(x)&=\lim_{x\to a}\sum_{i=0}^{n}a_ix^i\\ &=\sum_{i=0}^{n}\lim_{x\to a}a_ix^i\tag{summation rule of limit}\\ &=\sum_{i=0}^{n}a_ia^i\tag{lemma 1}\\ &=P(a)\\ &Q.E.D \end{align} $$
  2. In addition, Every rational function is also continuous Let the rational function be $R(x)=\frac{P_1(x)}{P_2(x)}$, given that $P_2(x)\neq 0$ $$ \begin{align} \lim_{x\to a}R(x)&=\lim_{x\to a}\frac{P_1(x)}{P_2(x)}\\ &=\frac{\lim_{x\to a}P_1(x)}{\lim_{x\to a}P_2(x)}\tag{reciprocal rule of limit}\\ &=\frac{P_1(a)}{P_2(a)}\tag{see proof of polynomial}\\ &=R(a)\\ &Q.E.D \end{align} $$

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