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Limits and Derivatives - Limits

Grade 11CBSE

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

πŸ”‘Concepts

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Limit of a Function: The limit of a function f(x)f(x) as xx approaches aa is the value LL that f(x)f(x) gets closer to as xx moves arbitrarily close to aa from both sides. Visually, imagine tracing the graph of f(x)f(x) with your finger from both the left and right toward the x-coordinate aa; the height (y-value) your finger approaches is the limit, denoted as lim⁑xβ†’af(x)=L\lim_{x \to a} f(x) = L.

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Left-Hand and Right-Hand Limits: A limit exists at x=ax = a if and only if the Left-Hand Limit (LHL), lim⁑xβ†’aβˆ’f(x)\lim_{x \to a^-} f(x), and the Right-Hand Limit (RHL), lim⁑xβ†’a+f(x)\lim_{x \to a^+} f(x), are equal. On a graph, this means the two ends of the curve must meet at the same point, even if there is a 'hole' exactly at x=ax = a.

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Indeterminate Forms: If substituting x=ax = a into the function results in an expression like 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}, it is called an indeterminate form. Visually, this often indicates a 'removable discontinuity' where the function is undefined at a single point but the surrounding curve remains continuous and points toward a specific value.

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Algebra of Limits: Limits distribute over basic operations. The limit of a sum is the sum of the limits, and the same applies to subtraction, multiplication, and division (provided the denominator's limit is not zero). This allows us to break down complex expressions into simpler parts to evaluate them step-by-step.

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Sandwich (Squeeze) Theorem: If we have three functions such that g(x)≀f(x)≀h(x)g(x) \le f(x) \le h(x) for all xx near aa, and both g(x)g(x) and h(x)h(x) approach the same limit LL at aa, then f(x)f(x) must also approach LL. Visually, the graph of f(x)f(x) is 'trapped' or 'squeezed' between the other two graphs, forcing it to pass through the same point at x=ax = a.

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Evaluation of Rational Limits: For rational functions, if direct substitution leads to 00\frac{0}{0}, we use algebraic techniques like factorization, rationalization of the numerator/denominator, or standard identities to simplify the expression and remove the factor causing the zero before re-evaluating.

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Trigonometric Limits: A fundamental concept is lim⁑xβ†’0sin⁑xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. Visually, this occurs because for very small values of xx (measured in radians), the length of the vertical segment sin⁑x\sin x in a unit circle becomes almost identical to the length of the arc xx, making their ratio approach 1.

πŸ“Formulae

lim⁑xβ†’a[f(x)Β±g(x)]=lim⁑xβ†’af(x)Β±lim⁑xβ†’ag(x)\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \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 (whereΒ 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{ (where } \lim_{x \to a} g(x) \neq 0)

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

lim⁑xβ†’0sin⁑xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

lim⁑xβ†’0cos⁑x=1\lim_{x \to 0} \cos x = 1

lim⁑xβ†’0tan⁑xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1

lim⁑xβ†’01βˆ’cos⁑xx=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 0

πŸ’‘Examples

Problem 1:

Evaluate the limit: 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_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} Since xβ†’3x \to 3 and xβ‰ 3x \neq 3, we can cancel the term (xβˆ’3)(x - 3): =lim⁑xβ†’3(x+3)=3+3=6= \lim_{x \to 3} (x + 3) = 3 + 3 = 6

Explanation:

Direct substitution results in 00\frac{0}{0}. By factorizing the numerator using the identity a2βˆ’b2=(aβˆ’b)(a+b)a^2 - b^2 = (a-b)(a+b), we can remove the indeterminate form and evaluate the limit.

Problem 2:

Evaluate the limit: lim⁑xβ†’0sin⁑4x7x\lim_{x \to 0} \frac{\sin 4x}{7x}

Solution:

lim⁑xβ†’0sin⁑4x7x=17β‹…lim⁑xβ†’0sin⁑4xx\lim_{x \to 0} \frac{\sin 4x}{7x} = \frac{1}{7} \cdot \lim_{x \to 0} \frac{\sin 4x}{x} Multiply and divide by 4 to match the standard formula: =47β‹…lim⁑xβ†’0sin⁑4x4x= \frac{4}{7} \cdot \lim_{x \to 0} \frac{\sin 4x}{4x} Let 4x=ΞΈ4x = \theta. As xβ†’0x \to 0, ΞΈβ†’0\theta \to 0: =47β‹…lim⁑θ→0sin⁑θθ=47β‹…1=47= \frac{4}{7} \cdot \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = \frac{4}{7} \cdot 1 = \frac{4}{7}

Explanation:

We use the standard trigonometric limit lim⁑xβ†’0sin⁑xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. By manipulating the expression to create a 4x4x in the denominator to match the sine argument, we can apply the formula.