RD Sharma Solutions for Class 11 Maths Chapter 29 - Free PDF Download
FAQs on RD Sharma Class 11 Maths Solutions Chapter 29 - Limits
1. How do RD Sharma Solutions for Class 11 Maths Chapter 29 (Limits) help in mastering the topic for the 2025-26 exams?
RD Sharma Solutions for Chapter 29 offer a comprehensive approach to mastering Limits. They provide a large number of solved problems that cover every concept in the CBSE syllabus, from the intuitive idea of limits to complex algebraic and trigonometric limits. By following the step-by-step methods, students can understand the correct procedure for solving questions, which is crucial for building a strong foundation in Calculus for Class 12.
2. What is the correct method for solving limit problems involving indeterminate forms like 0/0 in RD Sharma?
When a limit problem results in an indeterminate form like 0/0, the solutions in RD Sharma typically follow a structured approach:
- Step 1: First, confirm the indeterminate form by direct substitution.
- Step 2: Apply an appropriate algebraic method, such as factorisation of polynomials or rationalisation of expressions containing square roots.
- Step 3: Simplify the expression by cancelling out the common factors that cause the 0/0 form.
- Step 4: Substitute the limit value again into the simplified function to get the final answer.
3. How do the solutions for trigonometric limits in Chapter 29 apply standard limit formulas?
The solutions for trigonometric limits demonstrate how to manipulate expressions to match standard theorems. The goal is to isolate fundamental limit forms, such as lim (x→0) sin(x)/x = 1 and lim (x→0) (1-cos(x))/x = 0. The step-by-step solutions show how to use trigonometric identities, multiplication, and division to transform the original problem into a format where these standard results can be directly applied.
4. What is the key difference between a function's value f(a) and its limit as x approaches 'a'?
A function's value, f(a), is its exact output at the specific point x=a. In contrast, the limit of a function as x approaches 'a' describes the value that the function gets infinitesimally close to. A key insight is that the limit can exist even if the function is undefined at f(a). This concept is fundamental to understanding continuity and derivatives.
5. Why is it essential to check the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) for some functions?
Checking the LHL and RHL is crucial for functions that exhibit different behaviours on either side of a point, such as piecewise-defined functions, modulus functions, or greatest integer functions. A limit for a function at a point 'a' exists only if the value approached from the left (LHL) is exactly equal to the value approached from the right (RHL). If LHL ≠ RHL, the limit does not exist at that point.
6. How do the RD Sharma solutions typically solve problems involving limits at infinity?
For problems where x approaches infinity (∞), the standard technique shown in RD Sharma solutions is to divide both the numerator and the denominator by the highest power of x present in the denominator. This method effectively transforms the expression so that many terms become fractions with x in the denominator, which then approach zero as x approaches infinity. This simplifies the overall expression and makes the limit easy to evaluate.
7. When should direct substitution be used to find a limit according to the methods in RD Sharma?
Direct substitution is always the first step you should attempt when solving any limit problem. If substituting the value 'a' into the function f(x) yields a finite, defined number, then that number is the answer. You only need to proceed to other methods like factorisation or rationalisation if direct substitution results in an indeterminate form (e.g., 0/0, ∞/∞).
8. What are some key algebraic identities frequently used in solving limit problems in Chapter 29?
To simplify complex expressions, the solutions frequently use standard algebraic identities. The most common ones you must know are:
- (a² − b²) = (a − b)(a + b)
- (a³ − b³) = (a − b)(a² + ab + b²)
- (a³ + b³) = (a + b)(a² − ab + b²)

















