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NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives are provided here. This Class 11 NCERT syllabus introduces an important area of mathematics called calculus. This is a branch of mathematics which deal with the study of change in the value of a function as the points in the domain change. In this article, you will get limits and derivatives class 11 NCERT solutions that are prepared by highly qualified teachers keeping in mind latest syllabus of CBSE 2023, simple and provide step by step solution to each problem. Students can get all NCERT solutions at one place and practice them to develop indepth understanding of the concepts.
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In class 11 maths chapter 13 question answer, you will get questions related to all the above topics. This chapter is very important for both class 11 final examination and various competitive exams like JEE Main, JEE Advanced, VITEEE, BITSAT, etc. There are 43 questions in 2 exercises limits and derivatives class 11. The first exercise of this chapter deals with problems on limits and the second exercise deals with problems on derivatives. Students can find all NCERT solution for class 11 at one pace for hare and practice them.
Also Read :
Limits:
The limit of a function f(x) as x approaches a, denoted as lim(x→a) f(x), exists when both the left-hand limit (lim(x→a-) f(x)) and right-hand limit (lim(x→a+) f(x)) exist and are equal (lim(x→a-) f(x) = lim(x→a+) f(x)).
Left-Hand and Right-Hand Limits:
Left-Hand Limit: f(a-0) = lim(x→a-) f(x) = lim(h→0) f(a-h)
Right-Hand Limit: f(a+0) = lim(x→a+) f(x) = lim(h→0) f(a+h)
Properties of Limits:
If lim(x→a) f(x) and lim(x→a) g(x) both exist:
Sum Rule: lim(x→a) [f(x) ± g(x)] = lim(x→a) f(x) ± lim(x→a) g(x)
Constant Rule: lim(x→a) kf(x) = k lim(x→a) f(x)
Product Rule: lim(x→a) [f(x) * g(x)] = lim(x→a) f(x) * lim(x→a) g(x)
Quotient Rule: lim(x→a) [f(x)/g(x)] = [lim(x→a) f(x)] / [lim(x→a) g(x)] (provided lim(x→a) g(x) ≠ 0)
Standard Limits:
lim(x→a) [(xn - an)/(x - a)] = na(n-1)
lim(x→0) [(sin(x))/x] = 1
lim(x→0) [(tan(x))/x] = 1
lim(x→0) [(ax - 1)/x] = ln(a)
lim(x→0) [(ex - 1)/x] = 1
lim(x→0) [(ln(1+x))/x] = 1
Derivatives:
The derivative of a function f at a point x, denoted as f'(x), is defined as:
f'(x) = lim(h→0) [(f(x+h) - f(x))/h]
Properties of Derivatives:
If f(x) and g(x) have derivatives in a common domain:c
Sum Rule: [d/dx (f(x) + g(x))] = [d/dx (f(x))] + [d/dx (g(x))]
Difference Rule: [d/dx (f(x) - g(x))] = [d/dx (f(x))] - [d/dx (g(x))]
Product Rule: [d/dx (f(x) * g(x))] = [d/dx f(x)] * g(x) + f(x) * [d/dx g(x)]
Quotient Rule: [d/dx (f(x)/g(x))] = ([d/dx f(x)] * g(x) - f(x) * [d/dx g(x)]) / [g(x)]2
Standard Derivatives:
[d/dx (x^n)] = n * x(n-1)
[d/dx (sin(x))] = cos(x)
[d/dx (cos(x))] = -sin(x)
[d/dx (tan(x))] = sec2(x)
[d/dx (cot(x))] = -csc2(x)
[d/dx (sec(x))] = sec(x) * tan(x)
[d/dx (csc(x))] = -csc(x) * cot(x)
[d/dx (a^x)] = ax * ln(a)
[d/dx (e^x)] = ex
[d/dx (ln(x))] = 1/x
Free download NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives for CBSE Exam.
Limits and derivatives class 11 questions and answers - Exercise: 13.1
Question:1 Evaluate the following limits
Answer:
Answer
Question:6 Evaluate the following limits
Answer:
The limit
Lets put
since we have changed the function, its limit will also change,
so
So our function have became
Now As we know the property
Hence,
Question:8 Evaluate the following limits
Answer:
The limit
At x = 2 both numerator and denominator becomes zero, so lets factorise the function
Now we can put the limit directly, so
Question:10 Evaluate the following limits
Answer:
The limit
Here on directly putting limit , both numerator and the deniminator becomes zero so we factorize the function and then put the limit.
(Answer)
Question:11 Evaluate the following limits
Answer:
The limit:
Since Denominator is not zero on directly putting the limit, we can directly put the limits, so,
(Answer)
Question:12 Evaluate the following limits
Answer:
Here, since denominator becomes zero on putting the limit directly, so we first simplify the function and then put the limit,
(Answer)
Question:13 Evaluate the following limits
Answer:
The limit
Here on directly putting the limits, the function becomes form. so we try to make the function in the form of . so,
As
(Answer)
Question:14 Evaluate the following limits
Answer:
The limit,
On putting the limit directly, the function takes the zero by zero form So,we convert it in the form of .and then put the limit,
(Answer)
Question:16 Evaluate the following limits
Answer:
The limit
the function behaves well on directly putting the limit,so we put the limit directly. So.
(Answer)
Question:17 Evaluate the following limits
Answer:
The limit:
The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit
(Answer)
Question:18 Evaluate the following limits
Answer:
The function takes the form zero by zero when we put the limit directly in the function,. since function consist of sin function and cos function, we try to make the function in the form of as we know that it tends to 1 when x tends to 0.
So,
(Answer)
Question:19 Evaluate the following limits
Answer:
As function doesn't create any abnormality on putting the limit directly,we can put limit directly. So,
. (Answer)
Question:20 Evaluate the following limits
Answer:
The function takes the zero by zero form when we put the limit into the function directly, so we try to eliminate this case by simplifying the function. So
(Answer)
Question:21 Evaluate the following limits
Answer:
On putting the limit directly the function takes infinity by infinity form, So we simplify the function and then put the limit
(Answer)
Question:22 Evaluate the following limits
Answer:
The function takes zero by zero form when the limit is put directly, so we simplify the function and then put the limits,
So
Let's put
Since we are changing the variable, limit will also change.
as
So function in new variable becomes,
As we know tha property
(Answer)
Question:23 Find
Answer:
Given Function
Now,
Limit at x = 0 :
:
Hence limit at x = 0 is 3.
Limit at x = 1
Hence limit at x = 1 is 6.
Question:24 Find
Answer:
Limit at
Limit at
As we can see that Limit at is not equal to Limit at ,The limit of this function at x = 1 does not exists.
Question:25 Evaluate
Answer:
The right-hand Limit or Limit at
The left-hand limit or Limit at
Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.
Question:26 Evaluate
Answer:
The right-hand Limit or Limit at
The left-hand limit or Limit at
Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.
Question:27 Find
Answer:
The right-hand Limit or Limit at
The left-hand limit or Limit at
Since Left-hand limit and right-hand limit are equal, The limit of this function at x = 5 is 0.
Question:28 Suppose
f (x) = f (1) what are possible values of a and b?
Answer:
Given,
And
Since the limit exists,
left-hand limit = Right-hand limit = f(1).
Left-hand limit = f(1)
Right-hand limit
From both equations, we get that,
and
Hence the possible value of a and b are 0 and 4 respectively.
Question:29 Let a1, a2, . . ., an be fixed real numbers and define a function What is f (x) ? For some , compute l
Answer:
Given,
Now,
Hence
,
Now,
Hence
.
Question:30 If For what value (s) of a does exists ?
Answer:
Limit at x = a exists when the right-hand limit is equal to the left-hand limit. So,
Case 1: when a = 0
The right-hand Limit or Limit at
The left-hand limit or Limit at
Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.
Case 2: When a < 0
The right-hand Limit or Limit at
The left-hand limit or Limit at
Since LHL = RHL, Limit exists at x = a and is equal to a-1.
Case 3: When a > 0
The right-hand Limit or Limit at
The left-hand limit or Limit at
Since LHL = RHL, Limit exists at x = a and is equal to a+1
Hence,
The limit exists at all points except at x=0.
Question:32 If
Answer:
Given,
Case 1: Limit at x = 0
The right-hand Limit or Limit at
The left-hand limit or Limit at
Hence Limit will exist at x = 0 when m = n .
Case 2: Limit at x = 1
The right-hand Limit or Limit at
The left-hand limit or Limit at
Hence Limit at 1 exists at all integers.
Class 11 maths chapter 13 NCERT solutions - Exercise: 13.2
Question:1 Find the derivative of
Answer:
F(x)=
Now, As we know, The derivative of any function at x is
The derivative of f(x) at x = 10:
Question:2 Find the derivative of x at x = 1.
Answer:
Given
f(x)= x
Now, As we know, The derivative of any function at x is
The derivative of f(x) at x = 1:
(Answer)
Question:3 Find the derivative of 99x at x = l00.
Answer:
f(x)= 99x
Now, As we know, The derivative of any function at x is
The derivative of f(x) at x = 100:
Question:4 (i) Find the derivative of the following functions from first principle.
Answer:
Given
f(x)=
Now, As we know, The derivative of any function at x is
Question:4.(ii) Find the derivative of the following function from first principle.
Answer:
f(x)=
Now, As we know, The derivative of any function at x is
(Answer)
Question:4(iii) Find the derivative of the following functions from first principle.
Answer:
f(x)=
Now, As we know, The derivative of any function at x is
(Answer)
Question:4(iv) Find the derivative of the following functions from first principle.
Answer:
Given:
Now, As we know, The derivative of any function at x is
Question:5 For the function Prove that f '(1) =100 f '(0).
Answer:
As we know, the property,
applying that property we get
Now.
So,
Here
Hence Proved.
Question:6 Find the derivative of for some fixed real number a.
Answer:
Given
As we know, the property,
applying that property we get
Question:7(i) For some constants a and b, find the derivative of
Answer:
Given
As we know, the property,
and the property
applying that property we get
Question:7(ii) For some constants a and b, find the derivative of
Answer:
Given
As we know, the property,
and the property
applying those properties we get
Question:7(iii) For some constants a and b, find the derivative of
Answer:
Given,
Now As we know the quotient rule of derivative,
So applying this rule, we get
Hence
Question:8 Find the derivative of for some constant a.
Answer:
Given,
Now As we know the quotient rule of derivative,
So applying this rule, we get
Hence
Question:9(i) Find the derivative of
Answer:
Given:
As we know, the property,
and the property
applying that property we get
Question:9(ii) Find the derivative of
Answer:
Given.
As we know, the property,
and the property
applying that property we get
Question:9(iii) Find the derivative of
Answer:
Given
As we know, the property,
and the property
applying that property we get
Question:9(iv) Find the derivative of
Answer:
Given
As we know, the property,
and the property
applying that property we get
Question:9(v) Find the derivative of
Answer:
Given
As we know, the property,
and the property
applying that property we get
Question:9(vi) Find the derivative of
Answer:
Given
As we know the quotient rule of derivative:
and the property
So applying this rule, we get
Hence
Question:10 Find the derivative of from first principle.
Answer:
Given,
f(x)=
Now, As we know, The derivative of any function at x is
Question:11(i) Find the derivative of the following functions:
Answer:
Given,
f(x)=
Now, As we know the product rule of derivative,
So, applying the rule here,
Question:11(ii) Find the derivative of the following functions:
Answer:
Given
Now As we know the quotient rule of derivative,
So applying this rule, we get
Question:11 (iii) Find the derivative of the following functions:
Answer:
Given
As we know the property
Applying the property, we get
Question:11(iv) Find the derivative of the following functions:
Answer:
Given :
Now As we know the quotient rule of derivative,
So applying this rule, we get
Question:11(v) Find the derivative of the following functions:
Answer:
Given,
As we know the property
Applying the property,
Now As we know the quotient rule of derivative,
So applying this rule, we get
Question:11(vi) Find the derivative of the following functions:
Answer:
Given,
Now as we know the property
So, applying the property,
Question:11(vii) Find the derivative of the following functions:
Answer:
Given
As we know the property
Applying this property,
Limits and derivatives NCERT solutions - Miscellaneous Exercise
Question:1(i) Find the derivative of the following functions from first principle: -x
Answer:
Given.
f(x)=-x
Now, As we know, The derivative of any function at x is
Question:1(ii) Find the derivative of the following functions from first principle:
Answer:
Given.
f(x)=
Now, As we know, The derivative of any function at x is
Question:1(iii) Find the derivative of the following functions from first principle:
Answer:
Given.
Now, As we know, The derivative of any function at x is
Question:1(iv) Find the derivative of the following functions from first principle:
Answer:
Given.
Now, As we know, The derivative of any function at x is
Answer:
Given
f(x)= x + a
As we know, the property,
applying that property we get
Answer:
Given
As we know, the property,
applying that property we get
Answer:
Given,
Now,
As we know, the property,
and the property
applying that property we get
Answer:
Given,
Now, As we know the derivative of any function
Hence, The derivative of f(x) is
Hence Derivative of the function is
.
Answer:
Given,
Also can be written as
Now, As we know the derivative of any function
Hence, The derivative of f(x) is
Hence Derivative of the function is
Answer:
Given,
Now, As we know the derivative of any such function is given by
Hence, The derivative of f(x) is
Answer:
Given,
Now, As we know the derivative of any function
Hence, The derivative of f(x) is
Answer:
Given,
Now, As we know the derivative of any function
Hence, The derivative of f(x) is
Answer:
Given
As we know, the property,
and the property
applying that property we get
Answer:
Given
It can also be written as
Now,
As we know, the property,
and the property
applying that property we get
Answer:
Given
Now, As we know the chain rule of derivative,
And, the property,
Also the property
applying those properties we get,
Answer:
Given
Now, As we know the chain rule of derivative,
And the Multiplication property of derivative,
And, the property,
Also the property
Applying those properties we get,
Answer:
Given,
Now, As we know the chain rule of derivative,
Applying this property we get,
Answer:
Given,
the Multiplication property of derivative,
Applying the property
Hence derivative of the function is .
Answer:
Given,
Now, As we know the derivative of any function
Hence, The derivative of f(x) is
Answer:
Given
Also can be written as
which further can be written as
Now,
Answer:
Given,
which also can be written as
Now,
As we know the derivative of such function
So, The derivative of the function is,
Which can also be written as
.
Answer:
Given,
Now, As we know the chain rule of derivative,
And, the property,
Applying those properties, we get
Hence Derivative of the given function is
Answer:
Given Function
Now, As we know the derivative of any function of this type is:
Hence derivative of the given function will be:
Answer:
Given,
Now, As we know the derivative of any function
Hence the derivative of the given function is:
Answer:
Given
Now, As we know, the Multiplication property of derivative,
Hence derivative of the given function is:
Answer:
Given
Now, As we know the product rule of derivative,
The derivative of the given function is
Answer:
Given,
Now As we know the Multiplication property of derivative,(the product rule)
And also the property
Applying those properties we get,
Answer:
Given,
And the Multiplication property of derivative,
Also the property
Applying those properties we get,
Answer:
Given,
Now, As we know the derivative of any function
Also the property
Applying those properties,we get
Answer:
Given,
Now, As we know the derivative of any function
Now, As we know the derivative of any function
Hence the derivative of the given function is
Answer:
Given
Now, As we know the derivative of any function
Answer:
Given
Now, As we know the Multiplication property of derivative,
Also the property
Applying those properties we get,
the derivative of the given function is,
Answer:
Given,
Now, As we know the derivative of any function
Also chain rule of derivative,
Hence the derivative of the given function is
13.1 Introduction
13.2 Intuitive Idea of Derivatives
13.3 Limits
13.4 Limits of Trigonometric Functions
13.5 Derivatives
Interested students can practice class 11 maths ch 13 question answer using following exercises.
Limit of a function(f) at a point is the common value of the left-hand limits and right-hand limits if they coincide.
For a function 'f' and a real number 'a'. and may not be the same (One may be defined and not the other one).
For functions 'f' and 'g' the following properties holds:
chapter-1 | |
chapter-2 | |
chapter-3 | |
chapter-4 | |
chapter-5 | |
chapter-6 | |
chapter-7 | |
chapter-8 | |
chapter-9 | |
chapter-10 | |
chapter-11 | |
chapter-12 | |
chapter-13 | Limits and Derivatives |
chapter-14 | |
chapter-15 | |
chapter-16 |
Comprehensive coverage of concepts: The ch 13 maths class 11 provides a thorough and comprehensive coverage of the fundamental concepts of limits and derivatives, which are essential for understanding the principles of calculus.
Simple and easy language: The class 11 maths limits and derivatives is written in a simple and easy-to-understand language, making it accessible to students of all levels of mathematical aptitude.
Step-by-step explanations: The chapter provides step-by-step explanations of the concepts, with worked-out examples and exercises that help students to master the topics covered in the chapter.
Tip- Limits class 11 is more on the theory-based so you should understand every theorem and try to solve questions based on the theorem. Derivatives required more practice of problems. There are 30 questions in the miscellaneous exercise. You should solve miscellaneous exercises also to get command on this chapter with the help of NCERT solutions for class 11 maths chapter 13 limits and derivatives.
Effective way to study chapter 13 class 11 maths includes following points:
class11 maths ch13 includes following topics:
1. Introduction
2. Intuitive Idea of Derivatives
3. Limits
4. Limits of Trigonometric Functions
5. Derivatives
NCERT solutions for class 11 math ch 13 are helpful to the students in solving the NCERT problems as well building foundation for class 12th chapters such as differentiation, integration and many. These solutions are provided in a detailed manner which can be understood by an average student also.
Students can get the detailed NCERT solutions for class 11 maths here. they can practice these solutions to build concepts that lead confidence during exam and ultimately help to score well in the exam.
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