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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $P_2(x)\neq 0$ $$ \begin{align} \lim_{x\to a}R(x)&=\lim_{x\to a}\frac{P_1(x)}{P_2(x)}\\ &=\frac{\lim_{x\to a}P_1(x)}{\lim_{x\to a}P_2(x)}\tag{reciprocal rule of limit}\\ &=\frac{P_1(a)}{P_2(a)}\tag{see proof of polynomial}\\ &=R(a)\\ &Q.E.D \end{align} $$