CBSE Class 11 Math Chapter 4: Limits and Derivatives NCERT Solutions

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NCERT Solutions PDF Class 11 PDF

This chapter on Limits and Derivatives for CBSE Class 11 Mathematics (Maths II) introduces fundamental concepts of calculus. The NCERT Solutions provide detailed explanations and step-by-step problem-solving for exercises related to evaluating limits of various functions. It covers direct substitution, handling indeterminate forms like 0/0 using factorization and algebraic manipulation, and applying limit formulas. These solutions are designed to help students understand the core principles of limits, which are essential for grasping the concept of derivatives and further calculus topics. Practicing these problems will build a strong foundation for calculus and aid in exam preparation by clarifying common pitfalls and solution strategies.

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Quick info

BoardCBSE
ClassClass 11
Subjectगणित-II
Session2026
LanguageHindi
TypeNCERT Solutions
ChapterChapter 4

Chapter summary

Chapter 4, Limits and Derivatives, for Class 11 Mathematics NCERT Solutions focuses on understanding and calculating limits. The exercises cover direct substitution for continuous functions and algebraic methods like factorization to resolve indeterminate forms (0/0). Students will learn to apply standard limit formulas and manipulate expressions to find the limiting value of functions as the variable approaches a certain point. This chapter lays the groundwork for differential calculus.

Learning outcomes

  • Understand the concept of limits in calculus.
  • Evaluate limits of functions using direct substitution.
  • Apply algebraic techniques (factorization) to find limits for indeterminate forms.
  • Use the formula for the limit of x^n as x approaches a.
  • Solve problems involving limits of rational functions.

Topics covered

Paper topics

  • Introduction to Limits
  • Direct Substitution Method
  • Indeterminate Forms
  • Factorization Method for Limits
  • Limit Formula: \(\lim_{x\to a} \frac{x^n - a^n}{x - a}\)
  • Limits of Rational Functions
  • Evaluating Limits

Important topics

  • Evaluating Limits using Direct Substitution
  • Handling Indeterminate Forms (0/0)
  • Factorization Technique for Limits
  • Application of Limit Formulas

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Questions and Solutions

प्रश्न 1

सीमा ज्ञात कीजिए: \lim_{x \to 3} x + 3
Solution:

यह एक बहुपद फलन है, इसलिए हम सीमा का मान सीधे प्रतिस्थापित करके ज्ञात कर सकते हैं।

\lim_{x \to 3} x + 3 = 3 + 3 = 6

अतः, सीमा का मान 6 है।

प्रश्न 2

सीमा ज्ञात कीजिए: \lim_{x \to \pi} \left( x - \frac{22}{7} \right)
Solution:

यह एक बहुपद फलन है, इसलिए हम सीमा का मान सीधे प्रतिस्थापित करके ज्ञात कर सकते हैं।

\lim_{x \to \pi} \left( x - \frac{22}{7} \right) = \pi - \frac{22}{7}

अतः, सीमा का मान \(\pi - \frac{22}{7}\) है।

प्रश्न 3

सीमा ज्ञात कीजिए: \lim_{r\to 1} \pi r^2
Solution:

यह एक बहुपद फलन है, इसलिए हम सीमा का मान सीधे प्रतिस्थापित करके ज्ञात कर सकते हैं।

\lim_{r\to 1} \pi r^2 = \pi \times (1)^2 = \pi

अतः, सीमा का मान \(\pi\) है।

प्रश्न 4

सीमा ज्ञात कीजिए: \lim_{x \to 4} \frac{4x+3}{x-2}
Solution:

यह एक परिमेय फलन है। हम सीमा का मान सीधे प्रतिस्थापित करके ज्ञात कर सकते हैं क्योंकि हर शून्य नहीं होता है।

\lim_{x \to 4} \frac{4x+3}{x-2} = \frac{4 \times 4 + 3}{4-2} = \frac{16 + 3}{2} = \frac{19}{2}

अतः, सीमा का मान \(\frac{19}{2}\) है।

प्रश्न 5

सीमा ज्ञात कीजिए: \lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1}
Solution:

यह एक परिमेय फलन है। हम सीमा का मान सीधे प्रतिस्थापित करके ज्ञात कर सकते हैं क्योंकि हर शून्य नहीं होता है।

\lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1} = \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1} = \frac{1 + (-1) + 1}{-2} = \frac{1 - 1 + 1}{-2} = \frac{1}{-2} = -\frac{1}{2}

अतः, सीमा का मान \(-\frac{1}{2}\) है।

प्रश्न 6

सीमा ज्ञात कीजिए: \lim_{x\to 0} \frac{(x+1)^5-1}{x}

इस फलन में सीधे \(x=0\) रखने पर \(\frac{0}{0}\) रूप प्राप्त होता है, जो कि एक अनिर्धारित रूप है। इसलिए, हम एक उपयुक्त सूत्र का उपयोग करेंगे या चर का प्रतिस्थापन करेंगे।

Solution:

सीधे \(x=0\) रखने पर, हमें \(\(0+1)^5 - 1\) / \(0\) = \(1-1\)/\(0\) = \(\frac{0}{0}\) प्राप्त होता है। यह एक अनिर्धारित रूप है।

हम सूत्र \(\lim_{y\to a} \frac{y^n - a^n}{y - a} = na^{n-1}\) का उपयोग कर सकते हैं।

माना \(y = x+1\)। जब \(x \to 0\), तब \(y \to 0+1 = 1\)।

इसलिए, सीमा को इस प्रकार लिखा जा सकता है:

\lim_{x \to 0} \frac{(x+1)^5 - 1}{x} = \lim_{y \to 1} \frac{y^5 - 1}{y - 1}

यहाँ \(a=1\) और \(n=5\) है। सूत्र का प्रयोग करने पर:

\lim_{y \to 1} \frac{y^5 - 1^5}{y - 1} = 5 \times (1)^{5-1} = 5 \times 1^4 = 5

अतः, सीमा का मान 5 है।

प्रश्न 7

सीमा ज्ञात कीजिए: \lim_{x\to 2} \frac{3x^2-x-10}{x^2-4}

इस फलन में सीधे \(x=2\) रखने पर \(\frac{0}{0}\) रूप प्राप्त होता है, जो कि एक अनिर्धारित रूप है। इसलिए, हमें गुणनखंडन विधि का उपयोग करना होगा।

Solution:

सीधे \(x=2\) रखने पर, अंश \(3(2)^2 - 2 - 10 = 3(4) - 2 - 10 = 12 - 2 - 10 = 0\) और हर \(2^2 - 4 = 4 - 4 = 0\) होता है। यह \(\frac{0}{0}\) का अनिर्धारित रूप है।

हम अंश और हर का गुणनखंड करेंगे:

अंश: \(3x^2 - x - 10 = 3x^2 - 6x + 5x - 10 = 3x(x-2) + 5(x-2) = (x-2)(3x+5)\)

हर: \(x^2 - 4 = (x-2)(x+2)\)

अब सीमा को पुनः लिखें:

\lim_{x \to 2} \frac{(x-2)(3x+5)}{(x-2)(x+2)}

चूंकि \(x \to 2\), \(x \neq 2\), इसलिए हम \((x-2)\) को रद्द कर सकते हैं:

= \lim_{x \to 2} \frac{3x+5}{x+2}

अब हम सीधे \(x=2\) प्रतिस्थापित कर सकते हैं:

= \frac{3(2)+5}{2+2} = \frac{6+5}{4} = \frac{11}{4}

अतः, सीमा का मान \(\frac{11}{4}\) है।

Common mistakes

  • Incorrectly substituting values when direct substitution leads to an indeterminate form.
  • Errors in algebraic manipulation or factorization.
  • Forgetting to simplify expressions before substituting the limit value.
  • Misapplying limit formulas.

Revision tips

  • Review the conditions under which direct substitution is valid.
  • Practice factorizing polynomials to simplify rational functions before evaluating limits.
  • Memorize and correctly apply standard limit formulas.
  • Pay close attention to the indeterminate form (0/0) and the methods to resolve it.

Practice MCQs

Q1. What is the limit of the function \(f(x) = x + 3\) as \(x\) approaches 3?

Q2. When direct substitution results in \(\frac{0}{0}\), what is the typical next step?

Q3. What is the limit of \(\pi r^2\) as \(r\) approaches 1?

Q4. Evaluate \(\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}\).

Q5. The formula \(\lim_{x\to 0} \frac{x^n - a^n}{x - a} = na^{n-1}\) is used for which type of limit evaluation?

Frequently asked questions

What is the main focus of Chapter 4: Limits and Derivatives for Class 11 Maths?

This chapter focuses on understanding and calculating the limits of various functions. It covers methods like direct substitution and algebraic manipulation to handle indeterminate forms, laying the foundation for calculus.

How are limits evaluated in these NCERT Solutions?

The solutions demonstrate two primary methods: direct substitution for functions where it's applicable, and algebraic techniques like factorization for indeterminate forms (e.g., 0/0).

What does it mean if a limit results in an indeterminate form like \(\frac{0}{0}\)?

An indeterminate form like \(\frac{0}{0}\) means that direct substitution doesn't give a definitive answer. Further algebraic simplification or the use of specific limit formulas is required to find the actual limit.

Are there any specific formulas used in these solutions?

Yes, the solutions utilize formulas such as \(\lim_{x\to a} \frac{x^n - a^n}{x - a} = na^{n-1}\) to evaluate certain types of limits efficiently.

How do these solutions help with exam preparation?

These solutions provide clear, step-by-step explanations for each problem, helping students understand the concepts and methods required to solve limit problems, which is crucial for exams.

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