In our daily lives, we see the motion of vehicles, speedometers, and changing temperatures, which involve changing quantities. These changes are studied using limits and derivatives. So, what is a limit? A limit helps us understand the behaviour of a function as it approaches a particular value. A derivative tells us the rate of change of a function. For example, speed is the derivative of distance with respect to time. Derivatives are based on limits and form the foundation of calculus. The main purpose of these NCERT Notes is to provide students with an efficient study material from which they can revise the entire chapter.
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In these NCERT Class 11 Maths Notes, everything from definitions and properties to detailed notes, formulas, diagrams, and solved examples is fully covered by our subject matter experts at Careers360 to help the students understand the important concepts and feel confident about their studies. These NCERT Class 11 Maths Chapter 12 Notes are made in accordance with the latest NCERT syllabus while keeping it simple, well-structured and understandable. Find everything in one place – NCERT Books, Solutions, Syllabus, and Exemplar Problems with Solutions – in this NCERT article.
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Limits and Derivatives are fundamental concepts in calculus. A limit helps us understand the behaviour of a function as the input approaches a particular value, while a derivative measures how a function changes; in other words, it's the rate of change or slope of a curve at a point.
Let $f$ be a function defined in a domain which we take to be an interval, say, I. We shall study the concept of the limit of $f$ at a point $a$ in I.
We say $\lim\limits _{x \rightarrow a^{-}} f(x)$ is the expected value of $f$ at $x=a$ given the values of $f$ near to the left of $a$. This value is called the left-hand limit of $f$ at $a$.
We say $\lim\limits _{x \rightarrow a^{+}} f(x)$ is the expected value of $f$ at $x=a$ given the values of $f$ near to the right of $a$. This value is called the right-hand limit of $f$ at $a$.
If the right and left-hand limits coincide, we call the common value the limit of $f$ at $x=a$ and denote it by $\lim\limits _{x \rightarrow a} f(x)$.
If y=f(x) is a function where x=a, then it cannot take exact values, but closure to a, then limits are used to obtain a unique number when x tends to a or some value that is definite.
Mathematical representation: $\lim\limits _{x \rightarrow a} f(x)$
Left-Hand Limit:
If a function has values close to x=a on the left-hand side, then the unique value is given by:
$f\left(a^{-}\right)=\lim\limits _{x \rightarrow a^{-}} f(x)=\lim\limits _{h \rightarrow 0} f(a-h)$
Right-Hand Limit:
If a function that is given has values close to x=a on the right-hand, then the unique value is given by:
$f\left(a^{+}\right)=\lim\limits _{x \rightarrow a^{+}} f(x)=\lim\limits _{h \rightarrow 0} f(a+h)$
Theorem 1:
If f and g are two functions where both limits $\lim\limits _{x \rightarrow a} f(x)$ and $\lim\limits _{x \rightarrow a} g(x)$ exist.
i) The limit of the sum of two functions is equal to the limit of the individual sums of each function.
$\lim\limits _{x \rightarrow a}[f(x)+g(x)] = \lim\limits _{x \rightarrow a} f(x) + \lim\limits _{x \rightarrow a} g(x)$
ii) The limit of the difference between two functions is equal to the limit of the individual differences of each function.
$\lim\limits _{x \rightarrow a}[f(x)-g(x)] = \lim\limits _{x \rightarrow a} f(x) - \lim\limits _{x \rightarrow a} g(x)$
iii) The limit of the product of two functions is equal to the limit of the individual product of each function.
$\lim\limits _{x \rightarrow a}[f(x).g(x)] = \lim\limits _{x \rightarrow a} f(x) . \lim\limits _{x \rightarrow a} g(x)$
iv) The limit of the quotient of two functions is equal to the quotient of the limit of each function.
$\lim\limits _{x \rightarrow a}[\frac{f(x)}{g(x)}] = \frac{\lim\limits _{x \rightarrow a} f(x)}{\lim\limits _{x \rightarrow a} g(x)}$ [Provided $\lim\limits _{x \rightarrow a} g(x)\neq0$]
v) $\lim\limits _{x \rightarrow a} (λ.f(x))=λ.\lim\limits _{x \rightarrow a} f(x)$
If $f$ is a polynomial function, then $\lim\limits _{x \rightarrow a} f(x)$ exists and is given by
$\lim\limits _{x \rightarrow a} f(x)=f(a)$
An important limit, which is very useful and used in the sequel, is given below:
$\lim\limits _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}$
Remark: The above expression remains valid for any rational number provided that ' $a$ ' is positive.
Limits of trigonometric functions
To evaluate the limits of trigonometric functions, we shall make use of the following limits, which are given below:
(i) $\lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=1$
(ii) $\lim\limits _{x \rightarrow 0} \cos x=1$
(iii) $\lim\limits _{x \rightarrow 0} \sin x=0$
$\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}$
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
$\lim _{x \rightarrow 0} \frac{\tan x}{x}=1$
$\lim _{x \rightarrow 0} \frac{e^x-1}{x}=1$
$\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\ln a$
$\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$
1. For a polynomial function:
$
\begin{gathered}
\lim _{x \rightarrow a} x=a \\
\lim _{x \rightarrow a} x^n=a^n
\end{gathered}
$
2. For a rational function $f(x)=\frac{g(x)}{h(x)}$ :
$\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} \frac{g(x)}{h(x)}=\frac{\lim _{x \rightarrow a} g(x)}{\lim _{x \rightarrow a} h(x)}=\frac{g(a)}{h(a)}, \quad \text {if } h(a) \neq 0$
Theorem 2:
If $f(x)$ and $g(x)$ are two real-valued functions with the same domain such that
$f(x) \leq g(x)$
and if the limits $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ exist, then
$\lim _{x \rightarrow a} f(x) \leq \lim _{x \rightarrow a} g(x)$
Theorem 3: (Sandwich Theorem)
If $f(x), g(x)$, and $h(x)$ are real-valued functions with the same domain such that $f(x)≤g(x)≤h(x)$,
if $\lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x)$,
then $\lim _{x \rightarrow a} g(x)=l$
1. If $f$ is a real-valued function and $a$ belongs to its domain, the derivative of $f$ at $a$ is defined as:
$
f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}, \quad \text { if this limit exists. }
$
The derivative of $f$ at $a$ is denoted by $f^{\prime}(a)$.
2. Similarly, for a function $f(x)$, the derivative at $x$ is given by:
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}, \quad \text { if this limit exists. }$
The derivative of the sum of two functions is equal to the sum of the individual derivatives of each function.
$\frac{d}{d x}[f(x)+g(x)]=\frac{d}{d x} f(x)+\frac{d}{d x} g(x)$
The derivative of the difference of two functions is equal to the derivative of the individual difference of each function.
$\frac{d}{d x}[f(x)-g(x)]=\frac{d}{d x} f(x)-\frac{d}{d x} g(x)$
The derivative of the product of two functions is equal to the derivative of the individual product of each function.
$\frac{d}{d x}[f(x) \cdot g(x)]=\frac{d}{d x} f(x) \cdot g(x)+f(x) \cdot \frac{d}{d x} g(x)$
The derivative of the quotient of two functions is equal to the quotient of the derivatives of each function.
$\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{\frac{d}{d x} f(x) \cdot g(x)-f(x) \cdot \frac{d}{d x} g(x)}{[g(x)]^2}, \quad g(x) \neq 0$
If f(x) is said to be a polynomial function such that
$f(x)=a_n x^n+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\cdots+a_1 x+a_0$
Then the derivative of the function is given by:
$\frac{d}{d x} f(x)=n a_n x^{n-1}+(n-1) a_{n-1} x^{n-2}+(n-2) a_{n-2} x^{n-3}+\cdots+2 a_2 x+a_1$
1. $\frac{d}{d x}\left(x^n\right)=n x^{n-1}$
2. $\frac{d}{d x}(\sin x)=\cos x$
3. $\frac{d}{d x}(\cos x)=-\sin x$
4. $\frac{d}{d x}(\tan x)=\sec ^2 x$
5. $\frac{d}{d x}(\cot x)=-\operatorname{cosec}^2 x$
6. $\frac{d}{d x}(\sec x)=\sec x \cdot \tan x$
7. $\frac{d}{d x}(\operatorname{cosec} x)=-\operatorname{cosec} x \cdot \cot x$
8. $\frac{d}{d x}\left(a^x\right)=a^x \ln a$
9. $\frac{d}{d x}\left(e^x\right)=e^x$
10. $\frac{d}{d x}(\ln x)=\frac{1}{x}$
Given below are some previous year question answers of various examinations from the NCERT class 11 chapter 12 Limits and Derivatives:
Question 1: If $f(x)=1-x+x^2-x^3 \ldots-x^{99}+x^{100}$, then $\mathrm{f}^{\prime}(1)$ is equal to
Solution:
Given,
$\begin{aligned} & f(x)=1-x+x^2-x^3+\ldots-x^{99}+x^{100} \\ & f^{\prime}(x)=0-1+2 x-3 x^2+\ldots-99 x^{98}+100 x^{99} \\ & f^{\prime}(1)=-1+2-3+4-\ldots-99+100 \\ & =(2-1)+(4-3)+(6-5)+\ldots(100-99)=1+1+\ldots+1 \quad(50 \text { times })=50\end{aligned}$
Hence, the correct answer is 50.
Question 2: If $y=\frac{\sin (x+9)}{\cos x}$ then $\frac{d y}{d x}$ at $x=0$ is
Solution:
$\begin{aligned} & y=\frac{\sin (x+9)}{\cos x} \\ & \frac{d y}{d x}=\frac{\cos x \cos (x+9)-\sin (x+9)(-\sin x)}{\cos ^2 x} \\ & =\frac{\cos x \cos (x+9)+\sin (x+9) \sin x}{\cos ^2 x} \\ & =\frac{\cos (x+9-x)}{\cos ^2 x}=\frac{\cos 9^{\circ}}{\cos ^2 x} \\ & \left(\frac{d y}{d x}\right)_{x=0}=\frac{\cos 9^{\circ}}{(1)^2}=\cos 9^{\circ}\end{aligned}$
Hence, the correct answer is $\cos 9^{\circ}$.
Question 3: If $f(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots+\frac{x^{100}}{100}$ then $\mathrm{f}^{\prime}(1)$ is equal to:
Solution:
$\begin{aligned} & f(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots+\frac{x^{100}}{100} \\ & f^{\prime}(x)=0+1+\frac{2 x}{2}+\frac{3 x^2}{3}+\ldots+\frac{100 x^{99}}{100} \\ & f^{\prime}(x)=0+1+x+x^2+\ldots x^{99} \\ & f^{\prime}(1)=1+1+1+\ldots+1 \quad(100 \text { times })=100\end{aligned}$
Hence, the correct answer is 100.
All the links to chapter-wise notes for NCERT class 11 maths are given below:
After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better understanding of the concepts.
Students can also explore these well-structured, subject-wise Class 11 solutions for better understanding.
Students should always analyse the latest syllabus before making a study routine. The following links will help them check the syllabus and access some reference books.
Frequently Asked Questions (FAQs)
Left-hand limit: Value of f(x) as x approaches a point from the left.
Right-hand limit: Value of f(x) as x approaches a point from the right.
If Left-hand limit = Right-hand limit, the limit exists.
A limit helps in understanding the behaviour of a function near a point, while a derivative uses the concept of limits to measure the rate of change at that point.
Derivatives are used in calculating speed, growth rates, optimisation problems, and in physics, economics, and engineering.
Students can download the Free NCERT Class 11 Maths Chapter 12 Notes PDF Limits and Derivatives from the official Careers360 website.
A derivative measures the rate at which one quantity changes with respect to another. In simple terms, it represents the slope of a curve or the instantaneous rate of change of a function.
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