limr→ 1π r²

Exercise 12.1 · Q3

\lim_{r\to 1}\pi r^2

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Concept understanding — Limit Of Polynomial

What Happens to a Polynomial as $x$ Approaches a Number?

A limit answers a simple question about a polynomial: as $x$ gets closer and closer to some number $a$ , what value does the polynomial settle near?

The Intuition

Take $P(x) = 3x^2 - 2x + 1$ . What happens as $x$ gets really close to $2$ ?

At $x = 2$ , the polynomial gives $3(4) - 4 + 1 = 9$ .
At $x = 1.9$ , it gives about $8.63$ .
At $x = 2.1$ , it gives about $9.43$ .

The closer $x$ gets to $2$ , the closer $P(x)$ gets to $9$ . There is no drama — the polynomial just slides smoothly to that value.

Note

For any polynomial, $\lim_{x \to a} P(x) = P(a)$ . You can simply substitute the number.

The Precise Statement

\lim_{x \to a} P(x) = P(a)

Why this works. Using the algebra of limits, the limit of a sum is the sum of the limits, and the limit of a constant multiple is the constant times the limit. Since $\lim_{x \to a} x = a$ and $\lim_{x \to a} c = c$ , each term $c_k x^k$ tends to $c_k a^k$ . Adding the terms back together gives exactly $P(a)$ .

A Concrete Example

Find $\lim_{x \to 3} (2x^3 - 5x + 4)$ .

Step 1: Recognise it is a polynomial.

Step 2: Substitute $x = 3$ :

2(27) - 5(3) + 4 = 54 - 15 + 4 = 43

Tip

For polynomials, direct substitution is the only tool you need — no factoring, no rationalising. Just plug in and compute.

The One Trap: "But What If I Can't Plug In?"

You might wonder whether a polynomial can have a hole. It cannot — a polynomial is defined for every real number, so there is never a value you must avoid. The only time "just plug in" can fail is with a rational function (a polynomial divided by another polynomial), where the denominator might be zero. For a pure polynomial, the limit is always the value.

Why This Matters

Limits of polynomials are the foundation for:

Derivatives (the slope of a curve at a point),
Evaluating more complicated limits by simplifying to a polynomial first,
Solving problems in physics such as instantaneous velocity.

This is the simplest, most predictable limit in the whole chapter — everything else builds on it.

Limit of a Polynomial is one of the earliest results in the NCERT Class 11 Mathematics chapter on Limits and Derivatives, matching searches like "limits of polynomial functions formula" or "limits and derivatives important questions class 11". Because the direct-substitution rule is so reliable, it is a quick-scoring question type in both CBSE boards and JEE Main's calculus section.

Direct substitution in a polynomial limit: as $r \to 1$ , the expression $\pi r^2$ approaches $\pi(1)^2 = \pi$ .

Polynomials are the friendliest functions in calculus. They're continuous everywhere, which means the limit as you approach any point is simply the value at that point. No jumps, no holes, no drama.

The expression $\pi r^2$ is a polynomial in $r$ (specifically, a monomial of degree 2 with coefficient $\pi$ ). When we want to find its limit as $r \to 1$ , we're asking: what value does this expression get arbitrarily close to as $r$ gets arbitrarily close to 1?

Because polynomials are continuous, we can answer this question by direct substitution.

Solution

Recognize the function type. The expression $\pi r^2$ is a polynomial function of $r$ . Polynomials are continuous at every real number.
Apply the continuity property. For any continuous function $f$ and any point $a$ in its domain:

\lim_{x \to a} f(x) = f(a)

Substitute directly. Since $f(r) = \pi r^2$ is continuous at $r = 1$ :

\lim_{r \to 1} \pi r^2 = \pi (1)^2 = \pi \cdot 1 = \pi

Tip

For polynomial and rational functions (where the denominator is non-zero), always try direct substitution first. It works in the vast majority of basic limit problems.

✓Final answer

\boxed{\pi}

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Q.lim⁡r→1πr2\lim_{r\to 1}\pi r^2limr→1​πr2

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Q. $\lim_{r\to 1}\pi r^2$

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