NCERT Solutions for Class 11 Maths Chapter 13- Limits and Derivatives Exercise 13.1

The subject matter of Limits and Derivatives is a part of Mathematics but also has great relevance for the discipline of Physics. There are numerous calculations within these themes that are used both in Mathematics and Physics. This topic of Mathematics is very important for students who intend to study Physics at an advanced stage. The topic is covered for students in the NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1.The concepts of Physics are an integral part of everyday life and are applied almost everywhere. Especially in contemporary times, the majority of the utilities and gadgets humans use are a product of the application of the knowledge of Physics. It is important to be aware of the relevance of Mathematics as an academic discipline. There are numerous topics in Mathematics. A lot of calculations and concepts are taught as a part of these themes.The NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1 are great resources for the chapter on limits and derivatives.There are many subjects that use Mathematics when studied at a higher level. The concepts that have been discussed here, Limits and Derivatives are used in Physics, Chemistry, Biology and Economics. The chapter about Limits And Derivatives Class 11 Ex 13.1 discusses and explains the details and is informative about the implicit concepts and calculations. Exercise 13.1 Class 11 is an important portion of this chapter. The NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1 can help students learn and understand important themes, as well as clear up any doubts.

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In Mathematics, Derivatives are a part of Calculus. The topic of Calculus is not introduced to students in lower grades because a lot of prerequisite knowledge is required right at the start of the theme. It is introduced to students from a higher class after the necessary knowledge has been granted. The NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1 includes all of the prerequisite knowledge needed to grasp the concept of calculus.Examples in the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 start with the ideas of the speed and velocity of an object. It is evident that speed and velocity are part of the subject of Physics. The NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1, on the other hand, only cover the mathematical portion of it.It is very important to be well versed in the concepts and calculations of the topic in order to be able to efficiently and effectively solve the questions designated to them. The concepts and calculations of Derivatives can be accessed at the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 on the Extramarks website. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 are a reliable resource for learning.

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A PDF file is available for the students to download and access at their discretion. Students can refer to it in order to practise solving problems on a routine basis. Even if an Internet connection is not available, Class 11 Maths Chapter 13 Exercise 13.1 is readily available to students.Exercise 13.1 of Limits and Derivatives is based on the following topics:

As discussed above, the Topics of Limits and Derivatives have been based on Algebra and advanced Arithmetic. NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, and NCERT Solutions Class 4 are available on the Extramarks website to provide quick review facilities to students. NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions Class 9, and NCERT Solutions Class 10 are present on the Extramarks website for reference if the students require them. When students advance to class 11 and class 12, the Topics like Limits and Derivatives are introduced. The NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1 provide detailed information on the topics to help students understand them and clarify any doubts they may have.The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 are accessible through the Extramarks website.

When students learn Algebra, each level takes them a step higher and helps the student to understand the concept in detail. With the gradual and comprehensive increase in the complexity of calculations and formulae, students are able to accordingly enhance their capabilities, which in turn renders them capable of undertaking complex calculations. Consequently, this prepares them for concepts like Limits and Derivatives. NCERT Solutions are always available on the Extramarks website to assist students if they come across doubts and misconceptions. Class 11th Math Exercise 13.1 however, is produced specifically for the aforementioned topic. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 which are available on the Extramarks website provide extensive assistance to the students. They can always be accessed through the website for guidance, clarification, and a holistic understanding of the subject. Learners can also access the website to refer to previous chapters from previous classes in order to find the root of any misconceptions and to gain clarity with regard to recurring doubts. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 are also available in PDF format which can be downloaded by students. The content provided as a part of the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 is reliable, authentic, and uniform across the website.

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Access NCERT solutions for Class 11 Maths Chapter 13 – Limits and Derivatives

It is highly recommended that students access the NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 on the Extramarks website, as it is an asset that can be used by students anytime and anywhere. The topic has been discussed in detail, and the students have ample time to develop their understanding of the topic and its nuances. A student can access or download the study material and get assistance to learn the subject and revise it as many times as required. The NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1 have been designed with the NCERT guidelines in mind.This has been done in order to ensure that there are no discrepancies between the content of the solutions provided and the instruction imparted in schools. The NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1 are an invaluable resource designed to make life easier for students.Students can access the NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1 on the Extramarks website.NCERT Solutions for Class 11 Maths Chapters [Include Chapter wise Pages]

The NCERT has made earnest efforts in order to construct a comprehensive and well-structured framework for students across the country. Chapters compiled through extensive research have been compiled into a logical sequence in the form of a prescribed NCERT textbook for Class 11 to ensure that there is logical continuity in the delivery of content. The topics covered include Sets, Relations and Fractions, Linear Inequalities, Trigonometric Fractions, Conic Sections, Principle of Mathematical Inductions and Sequence of Series. Broad and intriguing themes such as Binomial Theorem, Straight Lines, Complex Numbers and Quadratic Equations, Permutations and Combinations, Introduction to Three-Dimensional Geometry, Limits and Derivatives, Mathematical Reasoning and Statistics and Probability have also been covered descriptively. Additionally, comprehensive and challenging assessments have been provided alongside these exercises for the purpose of evaluation. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 cover the theme of Limits and Derivatives. Students would benefit greatly from the NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1 during examinations.

NCERT Solution Class 11 Maths of Chapter 13 All Exercises

A number of exercises are included in the NCERT Solutions for Class 11 Maths, Chapter 13, Exercise 13.1.These solutions are available on the Extramarks website for the students’ convenience. Particularly, exercise 13.1 is based on Limits and Derivatives. In this chapter, there is an assortment of 73 questions that can be accessed and solved by a student. About 30 of the questions are of high difficulty.23 questions are simple questions that involve the application of appropriate formulas, and 20 of them are of medium difficulty. The topics that have been covered are available in NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1. This can be accessed on the Extramarks website.

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1

As is self-evident, the theme of Limits and Derivatives is inclusive of a number of diverse and sophisticated concepts and calculations. Algebra is a fundamental theme which is indispensable in order to adequately understand the theme of Limits. Polynomial and Trigonometric Functions are other concepts which have been deployed in the topic of Limits and Derivatives. Concepts that constitute the base of Limits and Derivatives have been rendered as part of the discipline of Mathematics in the lower classes, and students have to be skilled in order to take on the problems of Limits and Derivatives. It is advisable to access the NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1 for assistance and practice. The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 also gives students a chance to practise for board examinations, as it is built under the guidelines of NCERT.

Q.1 Evaluate the following limits in Exercises 1 to 22. 1.

$\lim_{x \to 3} (x + 3)$

$\lim_{x \to π} (x - \frac{22}{7})$

$\lim_{r \to 1} {πr}^{2}$

$\lim_{x \to 4} (\frac{4 x + 3}{x - 2})$

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

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

$\lim_{x \to 2} (\frac{3 x^{2} - x - 10}{x^{2} - 4})$

$\lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3}$

$\lim_{x \to 0} (\frac{ax + b}{cx + 1})$

10.

$\lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$

11.

$\lim_{x \to 1} (\frac{{ax}^{2} + bx + c}{{cx}^{2} + bx + a}), a + b + c \neq 0$

12.

$\lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

13.

$\lim_{x \to 0} \frac{sinax}{bx}$

14.

$\lim_{x \to 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

15.

$\lim_{x \to π} \frac{\sin (π - x)}{π (π - x)}$

16.

$\lim_{x \to 0} \frac{\cos x}{(π - x)}$

17.

$\lim_{x \to 0} \frac{\cos 2 x - 1}{cosx - 1}$

18.

$\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}$

19.

$\lim_{x \to 0} x \sec x$

20.

$\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}, a, b, a + b \neq 0$

21.

$\lim_{x \to 0} (cosec x - \cot x)$

22.

$\lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}}$

Ans.
1.

$\begin{array}{l} \lim_{x \to 3} (x + 3) = 3 + 3 \\ = 6 \end{array}$

$\lim_{x \to π} (x - \frac{22}{7}) = π - \frac{22}{7}$

$\begin{array}{l} \lim_{r \to 1} {πr}^{2} = π {(1)}^{2} \\ = π \end{array}$

$\begin{array}{l} \lim_{x \to 4} (\frac{4 x + 3}{x - 2}) = \frac{4 (4) + 3}{4 - 2} \\ = \frac{19}{2} \end{array}$

$\begin{array}{l} \lim_{x \to - 1} (\frac{x^{10} + x^{5} + 1}{x - 1}) = \frac{{(- 1)}^{10} + {(- 1)}^{5} + 1}{(- 1) - 1} \\ = \frac{1 - 1 + 1}{- 1 - 1} \\ = - \frac{1}{2} \end{array}$

$\begin{array}{l} \lim_{x \to 0} \frac{{(x + 1)}^{5} - 1}{x} \\ Let x + 1 = y \Rightarrow x = y - 1 \\ Then, \\ \lim_{y \to 1} \frac{y^{5} - 1}{y - 1} = \lim_{y \to 1} \frac{y^{5} - 1^{5}}{y - 1} \\ = 5 {(1)}^{5 - 1} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = {na}^{n - 1}] \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = 5 \\ ∴ \lim_{x \to 0} \frac{{(x + 1)}^{5} - 1}{x} = 5 . \end{array}$

$\begin{array}{l} \lim_{x \to 2} (\frac{3 x^{2} - x - 10}{x^{2} - 4}) = \lim_{x \to 2} {\frac{3 x^{2} - 6 x + 5 x - 10}{(x - 2) (x + 2)}} \\ = \lim_{x \to 2} {\frac{3 x (x - 2) + 5 (x - 2)}{(x - 2) (x + 2)}} \\ = \lim_{x \to 2} {\frac{(x - 2) (3 x + 5)}{(x - 2) (x + 2)}} \\ = \lim_{x \to 2} {\frac{(3 x + 5)}{(x + 2)}} \\ = \frac{3 (2) + 5}{2 + 2} \\ = \frac{11}{4} \end{array}$

$\begin{array}{l} \lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3} = \lim_{x \to 3} \frac{{(x^{2})}^{2} - 9^{2}}{2 x^{2} - 6 x + x - 3} \\ = \lim_{x \to 3} \frac{(x^{2} - 9) (x^{2} + 9)}{2 x (x - 3) + 1 (x - 3)} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \lim_{x \to 3} \frac{(x - 3) (x + 3) (x^{2} + 9)}{(x - 3) (2 x + 1)} \\ = \lim_{x \to 3} \frac{(x + 3) (x^{2} + 9)}{(2 x + 1)} \\ = \lim_{x \to 3} \frac{(3 + 3) (3^{2} + 9)}{(2 \times 3 + 1)} \\ = \frac{6 \times 18}{7} \\ = \frac{108}{7} \end{array}$

$\begin{array}{l} \lim_{x \to 0} (\frac{ax + b}{cx + 1}) = \frac{a (0) + b}{c (0) + 1} \\ = \frac{b}{1} \\ = b \end{array}$

10.

$\begin{array}{l} \lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} = \lim_{z \to 1} \frac{{(z^{\frac{1}{6}})}^{2} - 1}{z^{\frac{1}{6}} - 1} \\ = 2 (1^{2 - 1}) [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = {na}^{n - 1}] \\ = 2 \end{array}$

11.

$\begin{array}{l} \lim_{x \to 1} (\frac{{ax}^{2} + bx + c}{{cx}^{2} + bx + a}) = \frac{a {(1)}^{2} + b (1) + c}{c {(1)}^{2} + b (1) + a} \\ = \frac{a + b + c}{c + b + a} \\ = 1 \end{array}$

12.

$\begin{array}{l} \lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \to - 2} \frac{\frac{2 + x}{2 x}}{x + 2} \\ = \lim_{x \to - 2} \frac{x + 2}{2 x (x + 2)} \\ \lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \frac{1}{2 (- 2)} \\ = - \frac{1}{4} \end{array}$

13.

$\begin{array}{l} \lim_{x \to 0} \frac{\sin ax}{bx} = \frac{1}{b} \lim_{x \to 0} \frac{\sin ax}{ax} \times a \\ = \frac{a}{b} \lim_{ax \to 0} (\frac{\sin ax}{ax}) \\ = \frac{a}{b} \times 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = \frac{a}{b} \end{array}$

14.

$\begin{array}{l} \lim_{x \to 0} \frac{sinax}{sinbx} = \frac{a \lim_{ax \to 0} (\frac{sinax}{ax})}{b \lim_{bx \to 0} (\frac{sinbx}{bx})} \\ = \frac{a \times 1}{b \times 1} [∵ \lim_{x \to 0} \frac{sinx}{x} = 1] \\ = \frac{a}{b} \end{array}$

15.

$\begin{array}{l} \lim_{x \to π} \frac{\sin (π - x)}{π (π - x)} = \frac{1}{π} {\lim_{π - x \to 0} \frac{\sin (π - x)}{(π - x)}} \\ = \frac{1}{π} (1) \\ = \frac{1}{π} \end{array}$

16.

$\begin{array}{l} \lim_{x \to 0} \frac{cosx}{(π - x)} = \frac{\cos 0}{(π - 0)} \\ = \frac{1}{π} [∵ \cos 0 = 1] \end{array}$

17.

$\begin{array}{l} \lim_{x \to 0} \frac{\cos 2 x - 1}{cosx - 1} = \lim_{x \to 0} \frac{1 - 2 \sin^{2} x - 1}{1 - 2 \sin^{2} (\frac{x}{2}) - 1} \\ = \lim_{x \to 0} \frac{- 2 \sin^{2} x}{- 2 \sin^{2} (\frac{x}{2})} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \lim_{x \to 0} \frac{(\frac{\sin^{2} x}{x^{2}})}{{\frac{\sin^{2} (\frac{x}{2})}{4 {(\frac{x}{2})}^{2}}}} \\ = 4 \lim_{x \to 0} \frac{{(\frac{sinx}{x})}^{2}}{{\frac{\sin (\frac{x}{2})}{(\frac{x}{2})}}^{2}} \\ = 4 \frac{{(1)}^{2}}{{(1)}^{2}} \\ \lim_{x \to 0} \frac{\cos 2 x - 1}{cosx - 1} = 4 \end{array}$

18.

$\begin{array}{l} \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} = \lim_{x \to 0} \frac{x (a + cosx)}{b \sin x} \\ = \frac{\lim_{x \to 0} (a + \cos x)}{b \lim_{x \to 0} (\frac{\sin x}{x})} \\ = \frac{(a + \cos 0)}{b (1)} \\ \lim_{x \to 0} \frac{ax + x \cos x}{b \sin x} = \frac{a + 1}{b} \end{array}$

19.

$\begin{array}{l} \lim_{x \to 0} x \sec x = 0 . \sec 0 \\ = 0 \times 1 \\ = 0 \end{array}$

20.

$\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} = \lim_{x \to 0} \frac{x (\frac{a \sin ax}{ax} + b)}{x (a + \frac{b \sin bx}{bx})}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \lim_{x \to 0} \frac{(\frac{a \sin ax}{ax} + b)}{(a + \frac{b \sin bx}{bx})} \\ = \frac{(a \times 1 + b)}{(a + b \times 1)} \\ = \frac{a + b}{a + b} \\ \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} = 1 \end{array}$

21.

$\begin{array}{l} \lim_{x \to 0} (cosec x - \cot x) = \lim_{x \to 0} (\frac{1}{sinx} - \frac{\cos x}{sinx}) \\ = \lim_{x \to 0} (\frac{1 - \cos x}{sinx}) \\ = \lim_{x \to 0} (\frac{2 \sin^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}) \\ = \lim_{x \to 0} (\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}) \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \lim_{x \to 0} (\tan \frac{x}{2}) \\ = \tan 0 \\ \lim_{x \to 0} (cosec x - \cot x) = 0 \end{array}$

22.

$\begin{array}{l} Given : \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} \\ Let x - \frac{π}{2} = y \Rightarrow x = \frac{π}{2} + y \\ If x \to \frac{π}{2} then y \to 0 \\ ∴ \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} = \lim_{y \to 0} \frac{\tan 2 (\frac{π}{2} + y)}{\frac{π}{2} + y - \frac{π}{2}} \\ = \lim_{y \to 0} \frac{\tan (π + 2 y)}{y} \\ = \lim_{y \to 0} \frac{\tan 2 y}{y} [∵ \tan (π + x) = \tan x] \\ = 2 \lim_{y \to 0} \frac{\tan 2 y}{2 y} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = 2 \times 1 [\lim_{x \to 0} \frac{\tan x}{x} = 1] \\ \lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} = 2 \end{array}$

Q.2 $Find \lim_{x \to 0} f (x) and \lim_{x \to 1} f (x), where f (x) = {\begin{cases} 2 x + 3, x \leq 0 \\ 3 (x + 1), x > 0 \end{cases}$

Ans.

$\begin{array}{l} The given function is : f (x) = {\begin{cases} 2 x + 3, x \leq 0 \\ 3 (x + 1), x > 0 \end{cases} \\ At x = 0, \\ L . H . L . = \lim_{x \to 0^{-}} f (x) \\ = \lim_{x \to 0} (2 x + 3) \\ = 2 (0) + 3 \\ = 3 \\ R . H . L . = \lim_{x \to 0^{+}} f (x) \\ = \lim_{x \to 0} 3 (x + 1) \\ = 3 (0 + 1) \\ = 3 \\ ∴ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} f (x) = 3 \\ At x = 1, \\ L . H . L . = \lim_{x \to 1^{-}} f (x) \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \lim_{x \to 1} 3 (x + 1) \\ = 3 (1 + 1) \\ = 6 \\ R . H . L . = \lim_{x \to 1^{+}} f (x) \\ = \lim_{x \to 1} 3 (x + 1) \\ = 3 (1 + 1) \\ = 6 \\ ∴ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} f (x) = 6 \end{array}$

Q.3 $Find \lim_{x \to 1} f (x), where f (x) = {\begin{cases} x^{2} - 1, x \leq 1 \\ - x^{2} - 1, x > 1 \end{cases}$

Ans.

$\begin{array}{l} The given function is; f (x) = {\begin{cases} x^{2} - 1, x \leq 1 \\ - x^{2} - 1, x > 1 \end{cases} \\ At x = 1, \\ L . H . L . = \lim_{x \to 1^{-}} f (x) \\ = \underset{x \to 1}{im} (x^{2} - 1) \\ = {(1)}^{2} - 1 \\ = 0 \\ R . H . L . = \lim_{x \to 1^{+}} f (x) \\ = \lim_{x \to 1} (- x^{2} - 1) \\ = - {(1)}^{2} - 1 \\ = - 2 \\ ∴ \lim_{x \to 1^{-}} f (x) \lim_{x \to 1^{+}} f (x) \\ Thus, \lim_{x \to 1} f (x) does not exist . \end{array}$

Q.4 $Find \lim_{x \to 0} f (x), where f (x) = {\begin{cases} \frac{x}{| x |}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

Ans.

$\begin{array}{l} The given function is, \\ f (x) = {\begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases} \\ At x = 0, \\ L . H . L . = lim_{x \to 0^{-}} f (x) \\ = lim_{x \to 0^{-}} (\frac{x}{|x|}) \\ = lim_{x \to 0} (\frac{x}{- x}) [when x < 0, |x| = - x] \\ = - 1 \\ R . H . L . = lim_{x \to 0^{+}} f (x) \\ = lim_{x \to 0^{+}} (\frac{x}{|x|}) \\ = lim_{x \to 0} (\frac{x}{x}) [When x > 0, |x| = x] \\ = 1 \\ Since, lim_{x \to 0^{-}} f (x) lim_{x \to 0^{+}} f (x) \\ So, lim_{x \to 0} f (x) does not exist . \end{array}$

Q.5

$\begin{array}{l} Suppose f (x) = {\begin{cases} a + bx, x < 1 \\ 4, x = 1 \\ b - ax, x > 1 \end{cases} \\ and if \lim_{x \to 1} f (x) = f (1) what are possible values of a and b ? \end{array}$

Ans.

$\begin{array}{l} The given function is: f (x) = {\begin{cases} a + bx, x < 1 \\ 4, x = 1 \\ b - ax, x > 1 \end{cases} \\ For x = 1, \\ L.H.L. = \lim_{x \to 1^{-}} f (x) \\ = \lim_{x \to 1} (a + bx) \\ = a + b \\ R.H.L. = \lim_{x \to 1^{+}} f (x) \\ = \lim_{x \to 1} (b - ax) \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = b - a \\ and f (1) = 4 \\ According to given condition, \\ \lim_{x \to 1} f (x) = f (1) \\ \Rightarrow \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1) \\ \Rightarrow a + b = b - a = 4 \\ \Rightarrow a + b = 4 and b - a = 4 \\ Solving both equations, we get \\ a = 0 and b = 4 \\ Thus, the possible values of a and b are 0 and 4 respectively. \end{array}$

Q.6 $Find \lim_{x \to 5} f (x), where f (x) = | x | - 5$

Ans.

$\begin{array}{l} The given function is, \\ f (x) = | x | - 5 = {\begin{cases} - x - 5, x < 0 \\ - 5, x = 0 \\ x - 5, x > 0 \end{cases} \\ For x = 5, \\ L . H . L . = \lim_{x \to 5^{-}} f (x) \\ = \lim_{x \to 5} (x - 5) [When x>0, | x | = x] \end{array}$

$\begin{array}{l} = 5 - 5 \\ = 0 \\ R . H . L . = \lim_{x \to 5^{+}} f (x) \\ = \lim_{x \to 5} (x - 5) [When x>0, | x | = x] \\ = 5 - 5 \\ = 0 \\ ∴ \lim_{x \to 5^{-}} f (x) = \lim_{x \to 5^{+}} f (x) = \lim_{x \to 5} f (x) = 0 \\ Thus, \lim_{x \to 5} f (x) = 0 \end{array}$

Q.7 Let a₁, a₂, …, a_n be fixed real numbers and define a function f(x) = (x – a₁) (x – a₂)… (x – a_n).

$What is \lim_{x \to a_{1}} f (x) ? For some a \neq a_{1}, a_{2}, . .., a_{n}, compute \lim_{x \to a} f (x) .$

Ans.

$\begin{array}{l} The given function is: \\ f (x) = (x - a_{1}) (x - a_{2}) \dots (x - a_{n}) \\ ∴ \lim_{x \to a_{1}} f (x) = \lim_{x \to a_{1}} {(x - a_{1}) (x - a_{2}) \dots (x - a_{n})} \\ = {\lim_{x \to a_{1}} (x - a_{1})} {\lim_{x \to a_{1}} (x - a_{2})} \dots {\lim_{x \to a_{1}} (x - a_{n})} \\ = (a_{1} - a_{1}) (a_{1} - a_{2}) \dots (a_{1} - a_{n}) \\ = (0) (a_{1} - a_{2}) \dots (a_{1} - a_{n}) \\ = 0 \end{array}$ $And \lim_{x \to a} f (x)$

$= \lim_{x \to a} {(x - a_{1}) (x - a_{2}) \dots (x - a_{n})}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = {\lim_{x \to a} (x - a_{1})} {\lim_{x \to a} (x - a_{2})} \dots {\lim_{x \to a} (x - a_{n})} \\ = (a - a_{1}) (a - a_{2}) \dots (a - a_{n}) \end{array}$

Q.8

$\begin{array}{l} If f (x) = {\begin{cases} | x | + 1, & x < 0 \\ 0, & x = 0 \\ | x | - 1, & x > 0 \end{cases} \\ For what value (s) of a does \lim_{x \to a} f (x) exists ? \end{array}$

Ans.

$\begin{array}{l} The given function is: \\ f (x) = {\begin{cases} | x | + 1, x < 0 \\ 0, x = 0 \\ | x | - 1, x > 0 \end{cases} \\ For existence of \lim_{x \to a} f (x), \end{array}$

$\begin{array}{l} Case I: When a = 0 \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (|x| + 1) \\ = \lim_{x \to 0} (- x + 1) [When x < 0, |x| = - x] \\ = 0 + 1 \\ = 1 \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (|x| - 1) \\ = \lim_{x \to 0} (x - 1) [When x > 0, |x| = x] \\ = 0 - 1 \\ = - 1 \\ Here, \lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x) \\ ∴ \lim_{x \to 0} f (x) does not exist. \\ Case II : When a < 0 \\ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{-}} (|x| + 1) \\ = \lim_{x \to a} (- x + 1) [When x < 0, |x| = - x] \\ = - a + 1 \\ \lim_{x \to a^{+}} f (x) = \lim_{x \to a^{+}} (|x| + 1) \\ = \lim_{x \to a} (- x + 1) [When a < x < 0, |x| = - x] \\ = - a + 1 \\ ∴ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = - a + 1 \end{array}$

$\begin{array}{l} Thus, \lim_{x \to a} f (x) exists at x = a, where a<0. \\ Case III : When a > 0 \\ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{-}} (| x | - 1) \\ = \lim_{x \to a} (x - 1) [When 0 < x < a, | x | = x] \\ = a - 1 \\ \lim_{x \to a^{+}} f (x) = \lim_{x \to a^{+}} (| x | - 1) \\ = \lim_{x \to a} (x - 1) [When a < x < 0, | x | = x] \\ = a - 1 \\ ∴ \lim_{x \to a^{-}} f (x) = \lim_{x \to a^{+}} f (x) = a - 1 \\ Thus, \lim_{x \to a} f (x) exists at x = a, where a>0. \\ Therefore, \lim_{x \to a} f (x) exists for all a \neq 0. \end{array}$

Q.9

$\begin{array}{l} If the function f (x) satisfies \lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π, \\ evaluate \lim_{x \to 1} f (x) . \end{array}$

Ans.

$\begin{array}{l} G iven: \lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π \\ \lim_{x \to 1} {f (x) - 2} = \lim_{x \to 1} π (x^{2} - 1) \\ \Rightarrow \lim_{x \to 1} f (x) - \lim_{x \to 1} 2 = π (1^{2} - 1) \\ \Rightarrow \lim_{x \to 1} f (x) - \lim_{x \to 1} 2 = π (0) \end{array}$

$\begin{array}{l} \lim_{x \to 1} f (x) = \lim_{x \to 1} 2 \\ \Rightarrow \lim_{x \to 1} f (x) = 2 . \end{array}$

Q.10

$\begin{array}{l} If f (x) = {\begin{cases} {mx}^{2} + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ {nx}^{3} + m, & x > 0 \end{cases} . \\ For what integers m and n does both \lim_{x \to 0} f (x) and \lim_{x \to 1} f (x) exist ? \end{array}$

Ans.

$\begin{array}{l} We have, \\ f (x) = {\begin{cases} {mx}^{2} + n, x < 0 \\ nx + m, 0 \leq x \leq 1 \\ {nx}^{3} + m, x > 0 \end{cases} \\ For x = 0, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0} ({mx}^{2} + n) \\ = n \\ \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} ({nx}^{3} + m) \\ = m \\ Since, \lim_{x \to 0} f (x) exists. So, \\ \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) \\ \Rightarrow n = m \\ Therefore, \lim_{x \to 0} f (x) exists if m = n. \\ For x = 1, \end{array}$

$\begin{array}{l} \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1} (nx + m) \\ = n + m \\ \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} ({nx}^{3} + m) \\ = n + m \\ Since, \lim_{x \to 0} f (x) exists. So, \\ \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1} f (x) \\ Therefore, \lim_{x \to 1} f (x) exists for any value of m and n. \end{array}$

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FAQs (Frequently Asked Questions)

1. Is it necessary to study Limits and Derivatives? If Yes, Why?

It is necessary for Students of Class 11 to study Limits and Derivatives because it is a part of the syllabus set for class 11 and is therefore indispensable for securing good marks in the examinations. Moreover, practising Limits and Derivatives enhances the level of skill in a student and provides the capability to study subjects like Physics, Chemistry and Economics. There is also a significant portion of Biology that requires calculations which are made understandable with the aid of topics in Mathematics including Limits and Derivatives. It is recommended to use the Extramarks website to access the NCERT Solutions for Class 11 Maths, Chapter 13 Limits and Derivatives Exercise 13.1.

2. What are the chapters included in class 11 Mathematics?

There is an assortment of Topics which has been included in the chapters of Class 11. Some of the topics are Complex Numbers and Quadratic Equations, Introduction to Three-Dimensional Geometry, Relation and Fractions, Trigonometric Fractions, Principle of Mathematical Inductions, Binomial Theorem, Straight Lines, Permutations and Combinations, Limits and Derivatives, Statistics and Probability, Conic Sections, Mathematical Reasoning etc.The NCERT Solutions for Class 11 Maths, Chapter 13: Limits and Derivatives, Exercise 13.1 has thoroughly explained the various chapters in Class 11.

3. What are the chapters included in class 11 Mathematics?

There is an assortment of Topics which has been included in the chapters of Class 11. Some of the topics are Complex Numbers and Quadratic Equations, Introduction to Three-Dimensional Geometry, Relation and Fractions, Trigonometric Fractions, Principle of Mathematical Inductions, Binomial Theorem, Straight Lines, Permutations and Combinations, Limits and Derivatives, Statistics and Probability, Conic Sections, Mathematical Reasoning etc. The NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1 has explained in detail the assortments of chapters in class 11.

4. What is in chapter 13.1 class 11?

This particular chapter 13.1 discusses the Topic of Limits and Derivatives in detail and provides examples and problems to be solved. A student can find study material, questions, explanations, examples, etc., in the chapter. The NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1 have been particularly fashioned to help students as much as possible while they are attempting to acquire a flawless understanding of the topic. These solutions can be assessed through the Extramarks website, along with a PDF version that can be conveniently downloaded.

The NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1 have been developed to encourage and aid self-guided learning among students.

NCERT Solutions for Class 11 Maths Chapter 13- Limits and Derivatives Exercise 13.1

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1. Is it necessary to study Limits and Derivatives? If Yes, Why?

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4. What is in chapter 13.1 class 11?

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NCERT Solutions for Class 11 Maths Chapter 13- Limits and Derivatives Exercise 13.1

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FAQs (Frequently Asked Questions)

1. Is it necessary to study Limits and Derivatives? If Yes, Why?

2. What are the chapters included in class 11 Mathematics?

3. What are the chapters included in class 11 Mathematics?

4. What is in chapter 13.1 class 11?

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