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NCERT Solutions for Class 11 Maths Chapter 13: Limits and Derivatives Exercise 13.1- The NCERT Syllabus Class 11 Mathematics chapter limits and derivatives is an introduction to calculus. As the name of the chapter indicates, two main concepts are studied under this ex 13.1 class 11, that is limits and derivatives. The NCERT Solutions for exercise 13.1 Class 11 Maths chapter 13 deals with questions related to limits and algebra of limits. The questions of derivatives are covered in the next exercise.
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A few questions from the basic idea of limits are covered in the NCERT solutions for Class 11 Maths chapter 13 exercise 13.1. Also, exercise 13.1 Class 11 Maths has questions from limits of trigonometric functions. To solve many questions in the class 11 Maths chapter 13 exercise 13.1, the concepts of the algebra of limits are used. Limits of polynomial and rational functions are also covered in the Class 11th Maths chapter 13 exercise 13.1. A total of three exercises are given in the chapter including Class 11 Maths chapter 13 exercise 13.1.
**In the CBSE Syllabus for the academic year 2023-24, this chapter has been renumbered as Chapter 12.
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.
Some solved examples are given before exercise 13.1 Class 11 Maths. Before attempting the Class 11th Maths chapter 13 exercise 13.1 students are advised to go through the examples given in the NCERT book. Once examples are practised then it is important to solve the questions in Class 11 Maths chapter 13 exercise 13.1. Thirty-two questions and their explanations are given in NCERT solutions for Class 11 Maths chapter 13 exercise 13.1.
Also Read| Limits And Derivatives Class 11 Notes
Each and every question of exercise 13.1 Class 11 Maths are important to understand steps involved in solving basic problems of limits
Students may get the same or similar type of questions as in NCERT solutions for Class 11 Maths chapter 13 exercise 13.1 for Class and board exam
Limits is an important concept from JEE Main exam point of view.
Also see-
The conclusion is that the limit of g(x) as x tends to a does not exist.
The value of the given limit =1
The value of 1/cosx as x approaches 0 is 1
The given limits can be expressed as the product of limit (sinx/x) and 1/cosx as x approaches zero. Therefore limit of the function tanx/x as x tends to be zero=1
4 examples with their subquestions are given in the book
Thirty two
The main subject of interest of the exercise is the limits of different types of function(trigonometry and polynomial) and the algebra involved in limits
Yes the concepts of trigonometric limits and the unit limits, continuity and differentiability are important for JEE Main exam.
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