Q.Show that every polynomial function is continuous.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For polynomials, this follows directly from the limit laws.

Reasoning:

1. A polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$ is built from powers of $x$ multiplied by constants and added together.
2. For any real $a$, $\lim_{x \to a} x = a$, and by repeated application of the product law for limits, $\lim_{x \to a} x^k = a^k$ for any positive integer $k$.
3. Using the constant multiple and sum laws for limits: …

Why This Works: Continuity At A Point

Before we dive into the proof, let's get the intuition straight. A function is continuous at a point $x = a$ if three things hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

For polynomials, there's no division by zero, no square roots of negatives, no jumps — they're the "nice" functions of calculus. The trick is to build them from simpler pieces whose continuity we already know.

The last bullet is the engine of our proof. A polynomial is just a sum of terms like $c_n x^n$, and each $x^n$ is just $x$ multiplied by itself $n$ times. So if we can show $x$ is continuous, then $x^2$, $x^3$, ... are continuous by repeated multiplication, and then any constant times them is continuous, and finally the sum of all such terms is continuous.

Step-by-Step Proof

1. Start with the simplest building block: the identity function.

Let $f(x) = x$. For any real number $a$, we have:

So $f(x) = x$ is continuous at every $a \in \mathbb{R}$.

2. Show that $x^n$ is continuous for any positive integer $n$.

We use induction. Base case: $n=1$ is done above. Inductive step: assume $x^k$ is continuous at $a$. Then $x^{k+1} = x^k \cdot x$ is the product of two continuous functions ($x^k$ and $x$), so by the product rule for continuity, $x^{k+1}$ is continuous at $a$. By induction, $x^n$ is continuous for all $n \in \mathbb{N}$.

3. Multiply by a constant.

If $c$ is any real constant, then $c \cdot x^n$ is the product of the continuous constant function $g(x) = c$ and the continuous function $h(x) = x^n$. Since both are continuous at $a$, their product is continuous at $a$.

4. Add up the terms.

A general polynomial is:

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ …

Method: Proving an Entire Class of Functions Is Continuous Using the Algebra of Continuous Functions

This method applies whenever you must show that every function of a certain family (all polynomials, all rational functions, etc.) is continuous — not just verify one specific function at one specific point.

Steps

Step 1: Start from the smallest building blocks whose continuity is already established.

The two foundational facts are: a constant function $f(x)=c$ is continuous everywhere, and the identity function $f(x)=x$ is continuous everywhere (both follow directly from the definition of a limit).

Step 2: Invoke the algebra-of-continuous-functions theorem.

This lets you build up more complicated functions from the two basic blocks without ever returning to a fresh limit computation.

Step 3: Show every power $x^n$ is continuous.

Since $x^n$ is just $x$ multiplied by itself $n$ times, repeated application of the product rule (or a short induction) shows $x^n$ is continuous at every real number, for every positive integer $n$.

Step 4: Assemble the general element of the family. …

Common Mistakes

Mistake 1: "Proving" continuity by appealing to the graph looking smooth or unbroken.

Why it's wrong: a visual argument is not a proof — a rigorous "show that..." question requires the claim to be built from the definition of continuity via limit laws, not from "it looks continuous." Correct approach: always construct the argument from the algebra of continuous functions, starting from constants and the identity function.

Mistake 2: Treating the result as obvious and skipping the building-block steps. …