krit.club logo

Limits and Derivatives - Limits of polynomials and rational functions

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A polynomial function f(x)=anxn+an−1xn−1+⋯+a0f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 is continuous everywhere. Therefore, the limit as x→ax \to a is simply the value of the function at that point: lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a). This is known as direct substitution.

Graph of a polynomial function showing the limit at a point equals the function value.
•

A rational function is of the form R(x)=f(x)g(x)R(x) = \frac{f(x)}{g(x)}, where f(x)f(x) and g(x)g(x) are polynomials. If g(a)≠0g(a) \neq 0, the limit is found by direct substitution: lim⁡x→aR(x)=f(a)g(a)\lim_{x \to a} R(x) = \frac{f(a)}{g(a)}.

Graph of a rational function showing a vertical asymptote where the denominator is zero.
•

If direct substitution results in an indeterminate form like 00\frac{0}{0}, we must simplify the expression. For rational functions, this usually involves factoring the numerator and denominator to cancel the common factor (x−a)(x - a). For example, if f(x)=x2−a2x−af(x) = \frac{x^2 - a^2}{x - a}, the limit exists at x=ax = a even if the function is undefined there.

Linear graph with a hole representing a removable discontinuity in a rational function.
•

Algebra of limits: Limits distribute over addition, subtraction, multiplication, and division (provided the denominator limit is non-zero). This allows us to evaluate complex rational expressions by breaking them into simpler polynomial limits.

📐Formulae

lim⁡x→a[f(x)±g(x)]=lim⁡x→af(x)±lim⁡x→ag(x)\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) seeds

lim⁡x→a[f(x)⋅g(x)]=lim⁡x→af(x)⋅lim⁡x→ag(x)\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)

lim⁡x→af(x)g(x)=lim⁡x→af(x)lim⁡x→ag(x), provided lim⁡x→ag(x)≠0\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, \text{ provided } \lim_{x \to a} g(x) \neq 0

lim⁡x→axn−anx−a=nan−1\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}

💡Examples

Problem 1:

Evaluate lim⁡x→1(3x2+4x+5)\lim_{x \to 1} (3x^2 + 4x + 5).

Solution:

lim⁡x→1(3x2+4x+5)=3(1)2+4(1)+5=3+4+5=12\lim_{x \to 1} (3x^2 + 4x + 5) = 3(1)^2 + 4(1) + 5 = 3 + 4 + 5 = 12

Explanation:

Since the function is a polynomial, we use the direct substitution method.

Problem 2:

Find the limit: lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}.

Solution:

lim⁡x→2x2−4x−2=lim⁡x→2(x−2)(x+2)x−2=lim⁡x→2(x+2)=2+2=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 2 + 2 = 4

Explanation:

Direct substitution gives 00\frac{0}{0}. We factorize the numerator as (x−2)(x+2)(x-2)(x+2), cancel the common factor (x−2)(x-2), and then substitute x=2x=2.

Problem 3:

Evaluate lim⁡x→1x15−1x10−1\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}.

Solution:

lim⁡x→1x15−1x10−1=lim⁡x→1x15−115x−1x10−110x−1=15(1)1410(1)9=1510=32\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1} = \lim_{x \to 1} \frac{\frac{x^{15} - 1^{15}}{x - 1}}{\frac{x^{10} - 1^{10}}{x - 1}} = \frac{15(1)^{14}}{10(1)^9} = \frac{15}{10} = \frac{3}{2}

Explanation:

We divide both the numerator and denominator by (x−1)(x-1) to apply the standard formula lim⁡x→axn−anx−a=nan−1\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}.

Problem 4:

Evaluate the limit: lim⁡x→3x2−9x2−5x+6\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6}.

Graph of the simplified function showing the limit value 6 at x=3.

Solution:

Direct substitution gives 32−932−5(3)+6=00\frac{3^2 - 9}{3^2 - 5(3) + 6} = \frac{0}{0}, which is indeterminate. Factorize numerator and denominator: Numerator: x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3) Denominator: x2−5x+6=(x−3)(x−2)x^2 - 5x + 6 = (x - 3)(x - 2) So, lim⁡x→3(x−3)(x+3)(x−3)(x−2)\lim_{x \to 3} \frac{(x - 3)(x + 3)}{(x - 3)(x - 2)} Cancel the common factor (x−3)(x - 3) for x≠3x \neq 3: lim⁡x→3x+3x−2\lim_{x \to 3} \frac{x + 3}{x - 2} Now substitute x=3x = 3: 3+33−2=61=6\frac{3 + 3}{3 - 2} = \frac{6}{1} = 6.

Explanation:

Since substitution results in 00\frac{0}{0}, there is a common factor (x−3)(x-3) in both the numerator and denominator. Removing this 'hole' allows us to find the limit value.

Problem 5:

Find the value of lim⁡x→−1x3+1x+1\lim_{x \to -1} \frac{x^3 + 1}{x + 1}.

Parabolic graph showing the limit value of 3 as x approaches -1.

Solution:

Direct substitution gives (−1)3+1−1+1=00\frac{(-1)^3 + 1}{-1 + 1} = \frac{0}{0}. Use the algebraic identity a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2): x3+1=(x+1)(x2−x+1)x^3 + 1 = (x + 1)(x^2 - x + 1) So, lim⁡x→−1(x+1)(x2−x+1)x+1\lim_{x \to -1} \frac{(x + 1)(x^2 - x + 1)}{x + 1} Cancel (x+1)(x + 1): lim⁡x→−1(x2−x+1)\lim_{x \to -1} (x^2 - x + 1) Substitute x=−1x = -1: (−1)2−(−1)+1=1+1+1=3(-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3.

Explanation:

The limit represents the value the function approaches as xx gets closer to −1-1. By factoring the sum of cubes, we eliminate the term causing the division by zero.