Q.Examine whether the function $f$ given by $f(x) = x^2$ is continuous at $x = 0$.

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

1. Compute $f(0) = 0^2 = 0$.
2. Compute the limit: $\lim_{x \to 0} x^2 = 0$ (since $x^2$ is a polynomial, the limit equals the value at the point).
3. Since $\lim_{x \to 0} f(x) = f(0)$, the condition for continuity holds.

The Core Idea: Continuity at a Point

Continuity at a point means the function’s value and its limit agree — there’s no “jump” or “break” at that spot. Formally, $f$ is continuous at $x = a$ if:

For $f(x) = x^2$ at $x = 0$, we need to check two things:

1. Does $f(0)$ exist? Yes — $f(0) = 0^2 = 0$.
2. Does $\lim_{x \to 0} x^2$ exist and equal $0$?

The intuition: as $x$ gets closer to $0$, $x^2$ gets even closer to $0$ (since squaring a small number makes it smaller). There’s no sudden leap — the graph is a smooth parabola passing through the origin.

Step-by-Step Verification

1. Compute $f(0)$ directly.

Plugging $x = 0$ into $f(x) = x^2$ gives $f(0) = 0$. So the function is defined at the point.

2. Examine the left-hand limit ($x \to 0^-$).

If $x$ is negative but very close to $0$ (say $x = -0.1$), then $x^2 = 0.01$. As $x$ approaches $0$ from the left, $x^2$ approaches $0$. Formally:

3. Examine the right-hand limit ($x \to 0^+$).

If $x$ is positive and very close to $0$ (say $x = 0.1$), then $x^2 = 0.01$ again. As $x$ approaches $0$ from the right, $x^2$ also approaches $0$:

4. Compare the two one-sided limits.

Both are $0$, so the two-sided limit exists:

5. Check the equality condition.

We have $\lim_{x \to 0} f(x) = 0$ and $f(0) = 0$. Since they are equal, $f$ is continuous at $x = 0$.

Method: Checking Continuity at a Point

A function $f$ is continuous at $x=a$ only if all three of the following hold — checking just one or two is not a complete proof.

Steps

Step 1: Confirm $f(a)$ exists

Evaluate the function directly at the point.

Step 2: Confirm $\displaystyle\lim_{x\to a}f(x)$ exists

For most elementary functions this means checking that the left-hand limit and the right-hand limit are equal.

Step 3: Confirm the limit equals the function value

If all three conditions hold, $f$ is continuous at $a$; if any one fails, it is not.

Applying to this problem: for $f(x)=x^2$ at $x=0$, $f(0)=0$, and both one-sided limits of $x^2$ as $x\to0$ equal $0$ (since squaring a small number, positive or negative, gives an even smaller positive number) — all three conditions hold, so $f$ is continuous at $x=0$.

Common Mistakes

Mistake 1: Worrying that $x^2$ is always non-negative somehow affects whether the two one-sided limits match.

Why it's wrong: the sign of $f(x)$ near the point is irrelevant to continuity — what matters is that the VALUE $f(x)$ approaches from both sides is the same number, and $x^2\to0$ from both the negative and positive side identically. Correct approach: focus purely on the numerical value each one-sided limit approaches, not on the sign of intermediate values.

Mistake 2: Treating "it's a polynomial, so it's obviously continuous everywhere" as a substitute for the three-condition check at this specific point.

Why it's wrong: while true as a general theorem, a question that explicitly asks to "examine" continuity at a named point expects the explicit verification (computing $f(0)$ and both one-sided limits), not a citation of the general polynomial-continuity theorem alone. Correct approach: perform the full three-step check even when the conclusion is intuitively expected.