Q.Show that the function $f$ given by $f(x) = \begin{cases} x^3 + 3, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases}$ is not continuous at $x = 0$.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ iff $\lim_{x \to a} f(x) = f(a)$.

Step 1: Compute the limit as $x \to 0$. For $x \neq 0$, $f(x) = x^3 + 3$, so

Step 2: The function value at $x = 0$ is given directly: $f(0) = 1$. …

We need to check continuity at a single point. For a function $f$ to be continuous at $x = a$, three things must hold:

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

If any one of these fails, the function is discontinuous at that point. Here, the function is defined piecewise: it behaves like $x^3 + 3$ everywhere except at $x = 0$, where it is given a different value, $1$. That mismatch is the red flag.

Let’s check each condition.

1. $f(0)$ is defined.

   The definition says $f(0) = 1$. So condition 1 is satisfied.

2. Does $\lim_{x \to 0} f(x)$ exist?

   For $x \neq 0$, $f(x) = x^3 + 3$. As $x$ approaches $0$, $x^3$ approaches $0$, so $x^3 + 3$ approaches $3$.

   Since the function is given by the same expression $x^3 + 3$ for all $x \neq 0$, the left-hand limit and right-hand limit are both $3$.

   Therefore, $\displaystyle \lim_{x \to 0} f(x) = 3$.

3. Does $\lim_{x \to 0} f(x)$ equal $f(0)$? …

Method: Proving Discontinuity When a Point Is Redefined

This method is used when a piecewise function follows one formula almost everywhere but is deliberately assigned a different, isolated value at one point — the goal is to show the continuity definition fails there.

Steps

Step 1: Identify the "generic" formula and the "exceptional" value

Separate the function into the rule that applies for $x \neq a$ and the standalone value assigned at $x = a$ itself.

Step 2: Compute the limit using the generic formula

Since the one-sided behaviour near $a$ is governed entirely by the generic piece (the exceptional value at the single point $a$ doesn't affect the limit), compute

using ordinary substitution into the generic formula — the surrounding formula (often a polynomial) is typically continuous, so this limit exists and equals a direct substitution.

Step 3: Read off the function's actual value at the point …

Common Mistakes

Mistake 1: Concluding continuity just because $f(0)$ is defined

Why it's wrong: a function being defined at a point is only one of the three continuity conditions — it says nothing about whether the limit exists or matches that value. Correct approach: always compute $\lim_{x\to 0} f(x)$ separately and compare it with $f(0)$ before concluding anything.

Mistake 2: Substituting $x=0$ into the "$x\neq 0$" formula to find $f(0)$ …