Examine the continuity of the function $f(x) = x^3 + 2x^2 - 1$ at $x = 1$.

Examine the continuity of the function f(x) = x³ + – Class 12 Mathematics

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f(x) = x^3 + 2x^2 - 1

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Concept understanding — Continuity At A Point

Continuity at a Point

Imagine drawing the graph of a function and putting your pen down at $x = a$ . If the function is continuous there, you can draw straight through that point without lifting your pen — no jump, no hole, no break. That is the intuition; here is the precision.

The Three-Condition Test

For $f(x)$ to be continuous at $x = a$ , all three must hold. If even one fails, $f$ is discontinuous there.

Important

Continuity at $x = a$ requires:

$f(a)$ is defined,
$\displaystyle \lim_{x \to a} f(x)$ exists (left- and right-hand limits are equal),
$\displaystyle \lim_{x \to a} f(x) = f(a)$ .

Condition 1 says $a$ is in the domain — the pen must have somewhere to land. Condition 2 says the curve approaches a single value from both sides — no jump. Condition 3 says that common approach value actually matches the function's value at $a$ — no misplaced point.

Why All Three Are Needed

$f(x) = \dfrac{x^2 - 1}{x - 1}$ has $\lim_{x \to 1} f(x) = 2$ , yet $f(1)$ is undefined (zero denominator). Condition 1 fails, leaving a hole at $(1, 2)$ .

A piecewise function shows the opposite can be fine:

f(x) = \begin{cases} x + 1 & x < 2 \\ 3 & x = 2 \\ x + 1 & x > 2 \end{cases}

Here $f(2) = 3$ , both one-sided limits equal $3$ , and they match $f(2)$ — so all three hold and $f$ is continuous at $x = 2$ .

Common Pitfalls

Watch out

"Limit exists" does not mean "continuous." The hole example has a limit but no continuity — the limit must equal the function value.

Watch out

"Defined everywhere" does not mean "continuous." A piecewise function can have a value at every point and still jump. Always check the one-sided limits.

A Quick Check

Is $f(x) = |x|$ continuous at $x = 0$ ? $f(0) = 0$ , $\lim_{x \to 0}|x| = 0$ , and the two agree — yes, even though $|x|$ has a sharp corner. Continuity demands no break, not smoothness.

Takeaway: continuity at a point means the function value and the two one-sided limits all agree. Agreement $\Rightarrow$ the graph passes through unbroken; disagreement $\Rightarrow$ a discontinuity.

Continuity at a Point is the opening idea of the CBSE Class 12 Continuity and Differentiability chapter, and the three-condition test described here matches exactly what NCERT exercises and "continuity and differentiability class 12 important questions" expect students to apply. This concept is also foundational for JEE Main and NEET, where checking continuity is often the first step before testing differentiability of a function.

The function $f(x) = x^3 + 2x^2 - 1$ is a polynomial, and all polynomials are continuous at every real number. At $x = 1$ , the limit equals the function value: $\lim_{x \to 1} f(x) = f(1) = 2$ , so the function is continuous at $x = 1$ .

The Core Idea: Continuity at a Point

Before we touch a single calculation, let’s be clear on what “continuity at a point” actually means. A function $f$ is continuous at $x = a$ if three things happen together:

$f(a)$ exists (the function is defined at $a$ ).
$\lim_{x \to a} f(x)$ exists (the two-sided limit is a finite number).
The limit equals the function value: $\lim_{x \to a} f(x) = f(a)$ .

If any one of these fails, the function is discontinuous at that point. For most functions you meet in school, the tricky part is checking the limit — but here, we have a polynomial.

Tip

Polynomials are the “nice” functions of calculus. They have no holes, jumps, or vertical asymptotes. For any polynomial $p(x)$ , $\lim_{x \to a} p(x) = p(a)$ for every real $a$ . This is a theorem you can use directly in exams — no need to re-derive it each time.

So the problem reduces to: Is $f$ a polynomial? Yes. Then it’s continuous at $x = 1$ . But let’s verify it step by step anyway, because that’s how you build confidence.

Step-by-Step Verification

1. Check that $f(1)$ exists.

Plug $x = 1$ into the expression:

f(1) = (1)^3 + 2(1)^2 - 1 = 1 + 2 - 1 = 2.

The function is defined at $x = 1$ , and its value is $2$ . Condition 1 is satisfied.

2. Compute the two-sided limit as $x \to 1$ .

Since $f$ is a polynomial, we can evaluate the limit by direct substitution:

\lim_{x \to 1} f(x) = \lim_{x \to 1} (x^3 + 2x^2 - 1) = 1^3 + 2(1)^2 - 1 = 2.

The limit exists and equals $2$ . Condition 2 is satisfied.

Watch out

A common mistake is to think you always need to factor or simplify before taking a limit. That’s only necessary when direct substitution gives an indeterminate form like $\frac{0}{0}$ . Here, substitution works cleanly — don’t overcomplicate it.

3. Compare the limit and the function value.

We have:

\lim_{x \to 1} f(x) = 2 \quad \text{and} \quad f(1) = 2.

They are equal. Condition 3 is satisfied.

Since all three conditions hold, the function is continuous at $x = 1$ .

✓Final answer

f(x) = x^3 + 2x^2 - 1

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Q.Examine the continuity of the function $f(x) = x^3 + 2x^2 - 1$ at $x = 1$ .

Concept understanding — Continuity At A Point

Continuity at a Point

The Three-Condition Test

Why All Three Are Needed

Common Pitfalls

A Quick Check

The Core Idea: Continuity at a Point

Step-by-Step Verification

Method: Examining Continuity of a Polynomial at a Single Point Using the Three-Condition Test

Steps

Q.Examine the continuity of the function f(x)=x3+2x2−1f(x) = x^3 + 2x^2 - 1f(x)=x3+2x2−1 at x=1x = 1x=1.

The Core Idea: Continuity at a Point

Step-by-Step Verification

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Q.Examine the continuity of the function $f(x) = x^3 + 2x^2 - 1$ at $x = 1$ .

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