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Limits and Derivatives - Algebra of limits

Grade 11CBSE

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

πŸ”‘Concepts

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A limit describes the value a function f(x)f(x) approaches as the input xx gets closer and closer to a point aa.

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The limit exists only if the Left Hand Limit (LHL), lim⁑xβ†’aβˆ’f(x)\lim_{x \to a^-} f(x), is equal to the Right Hand Limit (RHL), lim⁑xβ†’a+f(x)\lim_{x \to a^+} f(x).

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The Algebra of Limits provides rules to calculate the limits of functions formed by the addition, subtraction, multiplication, and division of simpler functions.

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Indeterminate forms like 00\frac{0}{0} occur when direct substitution leads to undefined results; these require algebraic manipulation like factorization or rationalization.

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The 'Constant Multiple Rule' states that the limit of a constant times a function is the constant times the limit of the function.

πŸ“Formulae

lim⁑xβ†’a[f(x)+g(x)]=lim⁑xβ†’af(x)+lim⁑xβ†’ag(x)\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)

lim⁑xβ†’a[f(x)βˆ’g(x)]=lim⁑xβ†’af(x)βˆ’lim⁑xβ†’ag(x)\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)

lim⁑xβ†’a[f(x)β‹…g(x)]=lim⁑xβ†’af(x)β‹…lim⁑xβ†’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⁑xβ†’af(x)g(x)=lim⁑xβ†’af(x)lim⁑xβ†’ag(x),Β providedΒ lim⁑xβ†’ag(x)β‰ 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⁑xβ†’a[kβ‹…f(x)]=kβ‹…lim⁑xβ†’af(x)\lim_{x \to a} [k \cdot f(x)] = k \cdot \lim_{x \to a} f(x)

lim⁑xβ†’axnβˆ’anxβˆ’a=nanβˆ’1\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}

πŸ’‘Examples

Problem 1:

Evaluate the limit: lim⁑xβ†’2x2+5x+6x+2\lim_{x \to 2} \frac{x^2 + 5x + 6}{x + 2}

Solution:

lim⁑xβ†’2x2+5x+6x+2=22+5(2)+62+2=4+10+64=204=5\lim_{x \to 2} \frac{x^2 + 5x + 6}{x + 2} = \frac{2^2 + 5(2) + 6}{2 + 2} = \frac{4 + 10 + 6}{4} = \frac{20}{4} = 5

Explanation:

Since the denominator does not become zero at x=2x = 2, we can use the quotient rule and substitute the value of xx directly into the polynomial.

Problem 2:

Find the value of lim⁑xβ†’3x2βˆ’9xβˆ’3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Solution:

lim⁑xβ†’3x2βˆ’9xβˆ’3=lim⁑xβ†’3(xβˆ’3)(x+3)xβˆ’3=lim⁑xβ†’3(x+3)=3+3=6\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 3 + 3 = 6

Explanation:

Direct substitution gives 00\frac{0}{0}, which is indeterminate. We factorize the numerator using a2βˆ’b2=(aβˆ’b)(a+b)a^2 - b^2 = (a-b)(a+b) and cancel the common factor (xβˆ’3)(x-3) before applying the limit.

Problem 3:

Evaluate lim⁑xβ†’1[(x2+1)β‹…(3xβˆ’2)]\lim_{x \to 1} [ (x^2 + 1) \cdot (3x - 2) ]

Solution:

lim⁑xβ†’1(x2+1)β‹…lim⁑xβ†’1(3xβˆ’2)=(12+1)β‹…(3(1)βˆ’2)=2β‹…1=2\lim_{x \to 1} (x^2 + 1) \cdot \lim_{x \to 1} (3x - 2) = (1^2 + 1) \cdot (3(1) - 2) = 2 \cdot 1 = 2

Explanation:

Using the product rule of limits, we find the limits of the individual functions separately and then multiply them.