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

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

Edited By Vishal kumar | Updated on Nov 14, 2023 01:36 PM IST

## NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives Exercise 13.1- Download Free PDF

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.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

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.

## Question:1 Evaluate the following limits

$\lim_{x\rightarrow 3} x +3$

$\lim_{x\rightarrow 3} x +3$

$\Rightarrow \lim_{x\rightarrow 3} 3 +3$

$\Rightarrow 6$ Answer

Below you can find the solution:

$\lim_{x \rightarrow \pi } \left ( x - 22/7 \right )=\pi-\frac{22}{7}$

The limit

$\lim_{r \rightarrow 1} \pi r^2=\pi(1)^2=\pi$

Answer is $\pi$

The limit

$\lim_{x \rightarrow {4}} \frac{4x+3 }{x-2}$

$\Rightarrow \frac{4(4)+3 }{(4)-2}$

$\Rightarrow \frac{19 }{2}$ (Answer)

The limit

$\lim_{x \rightarrow {4}} \frac{x^{10}+ x^5 + 1 }{x-1}$

$\Rightarrow \frac{(-1)^{10}+ (-1)^5 + 1 }{(-1)-1}$

$\Rightarrow \frac{1-1+1}{-2}$

$\Rightarrow -\frac{1}{2}$ (Answer)

The limit

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

Lets put

$x+1=y$

since we have changed the function, its limit will also change,

so

$x\rightarrow 0,y\rightarrow 0+1=1$

So our function have became

$\lim_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }$

Now As we know the property

$\lim_{x \rightarrow 1 }\frac{ x^5 -a^n}{x-a }=na^{n-1}$

$\lim_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }=5(1)^5=5$

Hence,

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

The limit

$\lim_{x \rightarrow 2} \frac{3 x^2 - x -10 }{x^2 -4}$

$\Rightarrow \lim_{x \rightarrow 2} \frac{(x-2)(3x+5) }{(x-2)(x+2)}$

$\Rightarrow \lim_{x \rightarrow 2} \frac{(3x+5) }{(x+2)}$

$\Rightarrow \frac{(3(2)+5) }{((2)+2)}$

$\Rightarrow \frac{11 }{4}$ (Answer)

The limit

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

At x = 2 both numerator and denominator becomes zero, so lets factorise the function

$\lim_{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$

$\lim_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

Now we can put the limit directly, so

$\lim_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

$\Rightarrow \frac{((3)+3)((3)^2+9)}{(2(3)+1)}$

$\Rightarrow \frac{6\times18}{7}$

$\Rightarrow \frac{108}{7}$

The limit,

$\lim_{x \rightarrow 0 } \frac{ax +b}{cx+1}$

$\Rightarrow \frac{a(0) +b}{c(0)+1}$

$\Rightarrow b$ (Answer)

The limit

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

Here on directly putting limit , both numerator and the deniminator becomes zero so we factorize the function and then put the limit.

$\lim_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}=\lim_{z\rightarrow 1} \frac{z^{(1/6)^2}-1^2}{z^{1/6}-1}$

$=\lim_{z\rightarrow 1} \frac{(z^{(1/6)}-1)(z^{(1/6)}+1)}{z^{1/6}-1}$

$=1+1$

$=2$ (Answer)

The limit:

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

Since Denominator is not zero on directly putting the limit, we can directly put the limits, so,

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

$=\frac{a(1) ^2 +b(1) + c }{c(1)^2 + b(1) + a },$

$=\frac{a+b+c }{a+b+c },$

$=1$ (Answer)

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

Here, since denominator becomes zero on putting the limit directly, so we first simplify the function and then put the limit,

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

$=\lim_{x\rightarrow -2} \frac{\frac{x+2}{2x}}{x+2}$

$=\lim_{x\rightarrow -2} \frac{1}{2x}$

$= \frac{1}{2(-2)}$

$= -\frac{1}{4}$ (Answer)

The limit

$\lim_{x \rightarrow 0 } \frac{\sin ax }{bx }$

Here on directly putting the limits, the function becomes $\frac{0}{0}$ form. so we try to make the function in the form of $\frac{sinx}{x}$ . so,

$\lim_{x \rightarrow 0 } \frac{\sin ax }{bx }$

$=\lim_{x \rightarrow 0 } \frac{\sin ax(ax) }{bx(ax) }$

$=\lim_{x \rightarrow 0 } \frac{\sin ax }{ax }\frac{a}{b}$

As $\lim_{x\rightarrow 0}\frac{sinx}{x}=1$

$=1\cdot\frac{a}{b}$

$=\frac{a}{b}$ (Answer)

The limit,

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

On putting the limit directly, the function takes the zero by zero form So,we convert it in the form of $\frac{sina}{a}$ .and then put the limit,

$\Rightarrow \lim_{x\rightarrow {0}}\frac{\frac{sinax}{ax}}{\frac{sinbx}{bx}}\cdot\frac{ax}{bx}$

$=\frac{\lim_{ax\rightarrow {0}}\frac{sinax}{ax}}{\lim_{bx\rightarrow {0}}\frac{sinbx}{bx}}\cdot\frac{a}{b}$

$=\frac{a}{b}$ (Answer)

The limit

$\lim_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}$

$\Rightarrow \lim_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{ ( \pi -x)}\times\frac{1}{\pi}$

$= 1\times\frac{1}{\pi}$

$= \frac{1}{\pi}$ (Answer)

The limit

$\lim_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

the function behaves well on directly putting the limit,so we put the limit directly. So.

$\lim_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

$=\frac{\cos (0) }{\pi -(0) }$

$=\frac{1 }{\pi }$ (Answer)

The limit:

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

The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit

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

$\lim_{x\rightarrow 0}$ $\frac{-2(sin^2x)}{-2(sin^2(\frac{x}{2}))}$

$=\lim_{x\rightarrow 0}$ $\frac{\frac{(sin^2x)}{x^2}}{\frac{(sin^2(\frac{x}{2}))}{(\frac{x}{2})^2}}\times\frac{x^2}{(\frac{x}{2})^2}$

$=\frac{1^2}{1^2}\times 4$

$= 4$ (Answer)

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

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 $\frac{sinx}{x}$ as we know that it tends to 1 when x tends to 0.

So,

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

$=\frac{1}{b}\lim_{x \rightarrow 0}\frac{x(a+ \cos x) }{ \sin x }$

$=\frac{1}{b}\lim_{x \rightarrow 0}\frac{x }{ \sin x }\times(a+ \cos x)$

$=\frac{1}{b}\times1\times(a+ \cos (0))$

$=\frac{a+1}{b}$ (Answer)

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

As function doesn't create any abnormality on putting the limit directly,we can put limit directly. So,

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

$=(0) \sec (0)$

$=(0)\times 1$

$=0$ . (Answer)

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

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

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

$=\lim_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot ax+ bx }{ax + \frac{\sin bx}{bx}\cdot bx }$

$=\lim_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot a+ b }{a + \frac{\sin bx}{bx}\cdot b }$

$=\frac{1\cdot a+b}{a+1\cdot b}$

$=\frac{a+b}{a+ b}$

$=1$ (Answer)

$\lim_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

On putting the limit directly the function takes infinity by infinity form, So we simplify the function and then put the limit

$\lim_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

$=\lim_{x \rightarrow 0} \left (\frac{1}{sinx}-\frac{cosx}{sinx}\right )$

$=\lim_{x \rightarrow 0} \left (\frac{1-cosx}{sinx}\right )$

$=\lim_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{sinx}\right )$

$=\lim_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(\frac{x}{2})^2}{sinx} \right )$

$=\lim_{x \rightarrow 0} \frac{2}{4}\left (\frac{sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(x)}{sinx} \right )\cdot x$

$=\frac{2}{4}\times (1)^2\times0$

$=0$ (Answer)

$\lim_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }$

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

$y=x-\frac{\pi}{2}$

Since we are changing the variable, limit will also change.

as

$x\rightarrow \frac{\pi}{2},y=x-\frac{\pi}{2}\rightarrow \frac{\pi}{2}-\frac{\pi}{2}=0$

So function in new variable becomes,

$\lim_{y \rightarrow 0 } \frac{\tan 2(y+\frac{\pi}{2}) }{y+\pi/2- \pi /2 }$

$=\lim_{y \rightarrow 0 } \frac{\tan (2y+\pi) }{y }$

As we know tha property $tan(\pi+x)=tanx$

$=\lim_{y \rightarrow 0 } \frac{\tan (2y) }{y }$

$=\lim_{y \rightarrow 0 } \frac{sin2y}{2y}\cdot\frac{2}{cos2y}$

$=1\times 2$

$=2$ (Answer)

Given Function

$f (x) = \left\{\begin{matrix} 2x+3 & x \leq 0 \\ 3 ( x+1)& x > 0 \end{matrix}\right.$

Now,

Limit at x = 0 :

$at\:x=0^-$

: $\lim_{x\rightarrow{0^-}}f(x)=\lim_{x\rightarrow{0^-}}(2x+3)=2(0)+3=3$

$at\:x=0^+$

$\lim_{x\rightarrow{0^+}}f(x)=\lim_{x\rightarrow{0^+}}3(x+1)=3(0+1)=3$

Hence limit at x = 0 is 3.

Limit at x = 1

$at\:x=1^+$

$\lim_{x\rightarrow{1^+}}f(x)=\lim_{x\rightarrow{1^+}}3(x+1)=3(1+1)=6$

$at\:x=1^-$

$\lim_{x\rightarrow{1^-}}f(x)=\lim_{x\rightarrow{1^-}}3(x+1)=3(1+1)=6$

Hence limit at x = 1 is 6.

$\lim_{x \rightarrow 1} f (x ) , \: \:where \: \: f (x) = \left\{\begin{matrix} x^2 -1 & x \neq 0 \\ -x^2 -1 & x > 1 \end{matrix}\right.$

Limit at $x=1^+$

$\lim_{x \rightarrow 1^+} f (x ) = \lim_{x \rightarrow 1} (-x^2-1)=-(1)^2-1=-2$

Limit at $x=1^-$

$\lim_{x \rightarrow 1^-} f (x ) = \lim_{x \rightarrow 1} (x^2-1)=(1)^2-1=0$

As we can see that Limit at $x=1^+$ is not equal to Limit at $x=1^-$ ,The limit of this function at x = 1 does not exists.

$\lim_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or Limit at $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} \frac{|x|}{x}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} \frac{|x|}{x}=\frac{-x}{x}=-1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

$\lim_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or Limit at $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} \frac{x}{|x|}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} \frac{x}{|x|}=\frac{x}{-x}=-1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

$\lim_{x \rightarrow 5 } f (x) , where f( x) = |x| -5$

The right-hand Limit or Limit at $x=5^+$

$\lim_{x \rightarrow 5^+} f (x) = \lim_{x \rightarrow 5^+} |x|-5=5-5=0$

The left-hand limit or Limit at $x=5^-$

$\lim_{x \rightarrow 5^-} f (x) = \lim_{x \rightarrow 5^-}|x|-5=5-5=0$

Since Left-hand limit and right-hand limit are equal, The limit of this function at x = 5 is 0.

Question:28 Suppose

$f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$ f (x) = f (1) what are possible values of a and b?

Given,

$f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$

And

$\lim_{x\rightarrow 1} f(x)=f(1)$

Since the limit exists,

left-hand limit = Right-hand limit = f(1).

Left-hand limit = f(1)

$\lim_{x\rightarrow 1^-} f(x)= \lim_{x\rightarrow 1}(a+bx)=a+b(1)=a+b=4$

Right-hand limit

$\lim_{x\rightarrow 1^+} f(x)= \lim_{x\rightarrow 1}(b-ax)=b-a(1)=b-a=4$

From both equations, we get that,

$a=0$ and $b=4$

Hence the possible value of a and b are 0 and 4 respectively.

Given,

$f (x) = (x - a_1 ) (x - a_2 )...(x - a_n ) .$

Now,

$\\\lim_{x \rightarrow a _ 1 }f(x)=\lim_{x \rightarrow a _ 1 }[(x - a_1 ) (x - a_2 )...(x - a_n ) ]\\.=[\lim_{x \rightarrow a _ 1 }(x - a_1 )][\lim_{x \rightarrow a _ 1 }(x - a_2 )][\lim_{x \rightarrow a _ 1 }(x - a_n )] \\=0$

Hence

, $\lim_{x \rightarrow a _ 1 }f(x)=0$

Now,

$\lim_{ x \rightarrow a } f (x)=\lim_{ x \rightarrow a } (x-a_1)(x-a_2)...(x-a_n)$

$\lim_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$

Hence

$\lim_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$ .

$f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.$

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 $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} |x|-1=1-1=0$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} |x|+1=0+1=1$

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 $x=a^+$

$\lim_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|-1=a-1$

The left-hand limit or Limit at $x=a^-$

$\lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^-} |x|-1=a-1$

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 $x=a^+$

$\lim_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|+1=a+1$

The left-hand limit or Limit at $x=a^-$

$\lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^-} |x|+1=a+1$

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.

Given

$\lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi$

Now,

$\lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \frac{\lim_{x \rightarrow 1}(f (x)-2)}{\lim_{x \rightarrow 1}(x^2-1)}=\pi$

${\lim_{x \rightarrow 1}(f (x)-2)}=\pi{\lim_{x \rightarrow 1}(x^2-1)}$

${\lim_{x \rightarrow 1}(f (x)-2)}=\pi{(1-1)}$

${\lim_{x \rightarrow 1}(f (x)-2)}=0$

${\lim_{x \rightarrow 1}f (x)}=2$

Question:32 If

Given,

Case 1: Limit at x = 0

The right-hand Limit or Limit at $x=0^+$

$\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} nx+m=n(0)+m=m$

The left-hand limit or Limit at $x=0^-$

$\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} mx^2+n=m(0)^2+n=n$

Hence Limit will exist at x = 0 when m = n .

Case 2: Limit at x = 1

The right-hand Limit or Limit at $x=1^+$

$\lim_{x \rightarrow 1^+} f (x) = \lim_{x \rightarrow 1^+} nx^3+m=n(1)^3+m=n+m$

The left-hand limit or Limit at $x=1^-$

$\lim_{x \rightarrow 1^-} f (x) = \lim_{x \rightarrow 1^-} nx+m=n(1)+m=n+m$

Hence Limit at 1 exists at all integers.

## More About NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1

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

## Benefits of NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.1

• 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.

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## Key Features of NCERT Class 11 Maths Ex 13.1 Solution

• Step-by-Step Solutions: Ex 13.1 class 11 solutions are presented in a step-by-step manner, facilitating a clearer understanding of the intricacies involved in solving limit problems.
• Exam Significance: 11th class maths exercise 13.1 answers is deemed important for Class 11 board exams and is also considered significant for JEE Main exam preparation, particularly in establishing a strong foundation in understanding limits.
• Preparation Guidance: Students are recommended to review solved examples in the NCERT book before attempting class 11 maths chapter 13 exercise 13.1. This exercise comprises thirty-two questions accompanied by detailed explanations, ensuring a thorough understanding of the content.
• Conceptual Emphasis: The class 11 ex 13.1 predominantly deals with questions related to limits, establishing a conceptual foundation that becomes pivotal for understanding derivatives in subsequent exercises.

Also see-

## Subject Wise NCERT Exampler Solutions

1. what is the conclusion made if the right and left-hand limit of g(x) are different as x tends to ‘a’?

The conclusion is that the limit of g(x) as x tends to a does not exist.

2. Give the value of the limit of as x tends to 0.

The value of the given limit =1

3. Write the value of the limit of 1/cosx as x approaches 0

The value of 1/cosx as x approaches 0 is 1

4. What is the limit of the function tanx/x as x tends to zero

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

5. How many examples are solved under the topic limits in the NCERT Class 11 Maths book?

4 examples with their subquestions are given in the book

6. What is the number of questions given in the first exercise of the 11th Maths chapter 13?

Thirty two

7. What are the topic covered in the questions of exercise 13.1 Class 11 Maths?

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

8. Is the concepts of trigonometric limits important for JEE Main exam?

Yes the concepts of trigonometric limits and the unit limits, continuity and differentiability are important for JEE Main exam.

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