NCERT Solutions & Concepts: Limits and Derivatives Ex 13.1 (Class 11)

Understanding the Core Concepts of Limits

In algebra, we are used to evaluating a function $f(x)$ at exactly $x = a$ by computing $f(a)$. However, calculus introduces situations where a function might not be defined at $x = a$ (such as giving a $\frac{0}{0}$ form), yet we need to know what value the function approaches as $x$ gets arbitrarily close to $a$. This 'nearness' is captured by the limit: $\lim_{x \to a} f(x) = L$.

To evaluate limits algebraically in Exercise 13.1, we classify our problems into two scenarios:

1. Direct Substitution: If $f(x)$ is a polynomial or a rational function defined at $x = a$ (i.e., the denominator is non-zero), we simply plug in $x = a$ to find the limit.
2. Indeterminate Forms: If substituting $x = a$ yields an undefined expression like $\frac{0}{0}$, we must use algebraic manipulation to remove the vanishing factor before substituting.

Key Definitions and Standard Limit Formulas

Limit of a Function

The value $L$ that a function $f(x)$ approaches as the independent variable $x$ approaches a specified value $a$. Represented as $\lim_{x \to a} f(x) = L$.

Indeterminate Form

An undefined algebraic expression obtained during direct substitution, most commonly $\frac{0}{0}$ in algebraic limits, which requires further simplification.

Standard Algebraic Limit Theorem

For any positive integer $n$, the formula is: $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1}$.

Trigonometric Limit Identity

A fundamental trigonometric limit used in Exercise 13.1: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ (where $x$ is measured in radians).

Step-by-Step Strategy to Evaluate Limits

1. Step 1: Apply Direct Substitution — Substitute $x = a$ directly into the expression. If you obtain a real number, this real number is your final limit value.
2. Step 2: Check for Indeterminate Form — If direct substitution results in $\frac{0}{0}$, proceed to algebraic simplification. Do not leave $\frac{0}{0}$ as your final answer.
3. Step 3: Simplify Algebraically — Use techniques like Factorization (using algebraic identities like $a^2-b^2$ or $a^3-b^3$), Rationalization (if square roots are present), or the Standard Theorem formula $\lim_{x \to a} \frac{x^n-a^n}{x-a} = n a^{n-1}$.
4. Step 4: Cancel Common Factors and Re-substitute — Cancel out the common factor $(x - a)$ from both numerator and denominator, then substitute $x = a$ again to find the final numerical limit.

Fully Solved Examples of Class 11 Limits

• Example 1: Factorization Method Evaluate $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$. Step 1: Substitute $x = 2$ directly. This gives $\frac{2^3-8}{2^2-4} = \frac{0}{0}$ (Indeterminate form). Step 2: Factorize both the numerator and the denominator. Numerator: $x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)$ Denominator: $x^2 - 4 = (x-2)(x+2)$ Step 3: Rewrite the limit expression and cancel the common non-zero factor $(x-2)$: $\lim_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x+2}$ Step 4: Substitute $x = 2$ into the simplified expression: $\frac{2^2 + 2(2) + 4}{2+2} = \frac{12}{4} = 3$. Final Answer: The limit is 3.
• Example 2: Using the Standard Formula Evaluate $\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$. Step 1: Direct substitution yields $\frac{1-1}{1-1} = \frac{0}{0}$. Step 2: Divide both numerator and denominator by $(x-1)$ to utilize the standard limit formula: $\lim_{x \to 1} \frac{\frac{x^{15} - 1}{x - 1}}{\frac{x^{10} - 1}{x - 1}}$ Step 3: Apply the limit quotient rule and standard formula $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$: Numerator Limit: $\lim_{x \to 1} \frac{x^{15} - 1^{15}}{x - 1} = 15(1)^{14} = 15$ Denominator Limit: $\lim_{x \to 1} \frac{x^{10} - 1^{10}}{x - 1} = 10(1)^9 = 10$ Step 4: Divide the two values: $\frac{15}{10} = \frac{3}{2}$. Final Answer: The limit is 3/2.

YoLearn Exam Trap Alert!

Many CBSE students lose marks by writing mathematically incorrect steps.

1. Do not drop the 'lim' notation too early! Keep writing $\lim_{x \to a}$ in front of your expression in every single line until you actually substitute the value of $a$ into the equation.
2. The L'Hopital's Rule Trap: While L'Hopital's Rule is highly popular in JEE preparation for quickly evaluating $\frac{0}{0}$ limits, using it directly in CBSE Class 11 home exams might result in zero marks. CBSE board examiners look for formal algebraic methods (factorization, rationalization, or NCERT standard limit identities). Stick to NCERT methods to secure full step-marks!

Practice Questions with Solutions

• Q: Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$. A: Step 1: Check by substituting $x = 3$. This gives $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$ (Indeterminate form). Step 2: Factorize the numerator using the identity $a^2 - b^2 = (a-b)(a+b)$: $x^2 - 9 = (x - 3)(x + 3)$. Step 3: Rewrite the limit expression and cancel $(x-3)$: $\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3)$. Step 4: Now, perform direct substitution: $3 + 3 = 6$. Final answer: 6
• Q: Evaluate $\lim_{x \to 0} \frac{\sin 4x}{3x}$. A: Step 1: Direct substitution gives $\frac{\sin 0}{0} = \frac{0}{0}$ (Indeterminate form). Step 2: We must use the standard trigonometric limit: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Step 3: Manipulate the denominator to match the angle of the sine term ($4x$). Multiply and divide the expression by $4$: $\lim_{x \to 0} \frac{\sin 4x}{3x} = \lim_{x \to 0} \left( \frac{\sin 4x}{4x} \cdot \frac{4}{3} \right)$. Step 4: Since $x \to 0$, it implies $4x \to 0$. Apply the standard limit: $\frac{4}{3} \cdot \lim_{4x \to 0} \frac{\sin 4x}{4x} = \frac{4}{3} \cdot 1 = \frac{4}{3}$. Final answer: 4/3
• Q: Evaluate $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$. A: Step 1: Direct substitution of $x = -2$ gives $\frac{\frac{1}{-2} + \frac{1}{2}}{-2 + 2} = \frac{0}{0}$ (Indeterminate form). Step 2: Simplify the numerator by taking the common denominator: $\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x}$. Step 3: Rewrite the original limit expression using this simplification: $\lim_{x \to -2} \frac{\frac{x + 2}{2x}}{x + 2} = \lim_{x \to -2} \frac{x + 2}{2x(x + 2)}$. Step 4: Cancel the common term $(x + 2)$ and substitute $x = -2$: $\lim_{x \to -2} \frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}$. Final answer: -1/4
• Q: Evaluate $\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}$. A: Step 1: Direct substitution of $x = 2$ gives $\frac{3(4) - 2 - 10}{4 - 4} = \frac{0}{0}$ (Indeterminate form). Step 2: Factorize the quadratic trinomial in the numerator: $3x^2 - x - 10 = 3x^2 - 6x + 5x - 10 = 3x(x - 2) + 5(x - 2) = (3x + 5)(x - 2)$. Step 3: Factorize the denominator: $x^2 - 4 = (x - 2)(x + 2)$. Step 4: Cancel the common factor $(x - 2)$ and evaluate the limit: $\lim_{x \to 2} \frac{(3x + 5)(x - 2)}{(x + 2)(x - 2)} = \lim_{x \to 2} \frac{3x + 5}{x + 2} = \frac{3(2) + 5}{2 + 2} = \frac{11}{4}$. Final answer: 11/4

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