Home Class 11 Applied Maths Chapter 7: Calculus Exercise 7.1: Limits
Chapter 7 • Unit III

Exercise 7.1: Limits

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

7.1 Limits: Foundations, Left/Right Hand Bounds, and Indeterminate Forms

Calculus is the mathematical language of instantaneous change, continuous compounding, and economic marginality. The limit is the foundational gateway to calculus: it formalizes the behavior of a function near a point, rather than its value at the point itself, resolving apparent indeterminacies like \(\frac{0}{0}\).

Fundamental Definitions & Algebraic Limit Theorems

  • Intuitive Definition of a Limit: We write: \[ \lim_{x \to c} f(x) = L \] if the values of \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(c\) (from either side), but \(x \neq c\).
  • Existence Theorem (LHL = RHL): The two-sided limit exists if and only if both the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) exist and are equal: \[ \lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L \]
  • Algebra of Limits: If \(\lim_{x \to c} f(x) = L\) and \(\lim_{x \to c} g(x) = M\):
    1. \(\lim [f(x) \pm g(x)] = L \pm M\)
    2. \(\lim [k \cdot f(x)] = k \cdot L\) (for any constant \(k\))
    3. \(\lim [f(x) \cdot g(x)] = L \cdot M\)
    4. \(\lim \left[ \frac{f(x)}{g(x)} \right] = \frac{L}{M} \quad (\text{provided } M \neq 0)\)
    5. \(\lim [f(x)]^n = L^n\)
  • Standard Algebraic Limit Formula: For any real number \(a > 0\) and any rational number \(n\): \[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \]
  • Limits at Infinity (Asymptotic Long-Run Behavior): For rational functions \(\frac{P(x)}{Q(x)}\), divide numerator and denominator by the highest power of \(x\): \[ \lim_{x \to \infty} \frac{1}{x^k} = 0 \quad (\text{for } k > 0) \]
  • The Continuous Compounding Euler Limit: \[ \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e \approx 2.71828 \implies \lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e \] In continuous compound interest: \(A = \lim_{m \to \infty} P \left(1 + \frac{r}{m}\right)^{mt} = P e^{rt}\).

Systematic Techniques for Evaluating Indeterminate Limits \(\left(\frac{0}{0}\right)\)

Method Applicable Scenario Algebraic Strategy
Direct Substitution When \(f(c)\) evaluates to a definite real number Simply evaluate \(f(c)\)
Factorization Polynomial fractions yielding \(\frac{0}{0}\) Factor out and cancel the vanishing factor \((x - c)\)
Rationalization Expressions involving square roots yielding \(\frac{0}{0}\) Multiply numerator and denominator by the algebraic conjugate
Standard Formula Expressions resembling \(\frac{x^n - a^n}{x^m - a^m}\) Apply \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\)
Highest Power Division Limits as \(x \to \infty\) of rational expressions Divide numerator and denominator by highest power of \(x\)

Worked Example: Evaluating Algebraic and Rational Limits

Evaluate the following limits:

  1. \(\displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\)
  2. \(\displaystyle \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\)
  3. \(\displaystyle \lim_{x \to a} \frac{x^{5/2} - a^{5/2}}{x^{1/2} - a^{1/2}}\)

Solution:

  1. Factorization: Substituting \(x=2\) gives \(\frac{0}{0}\). \[ \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} = \frac{2^2 + 2(2) + 4}{2 + 2} = \frac{12}{4} = 3. \]
  2. Rationalization: Multiply numerator and denominator by \(\sqrt{1 + x} + 1\): \[ \lim_{x \to 0} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)} = \lim_{x \to 0} \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)} \] \[ = \lim_{x \to 0} \frac{x}{x(\sqrt{1 + x} + 1)} = \lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}. \]
  3. Using Standard Theorem: Divide numerator and denominator by \(x - a\): \[ \lim_{x \to a} \frac{\frac{x^{5/2} - a^{5/2}}{x - a}}{\frac{x^{1/2} - a^{1/2}}{x - a}} = \frac{\frac{5}{2} a^{\frac{5}{2} - 1}}{\frac{1}{2} a^{\frac{1}{2} - 1}} = \frac{\frac{5}{2} a^{3/2}}{\frac{1}{2} a^{-1/2}} = 5 a^{\frac{3}{2} - (-\frac{1}{2})} = 5 a^2. \]
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