Limits and Derivatives
Introduction to Limits and Derivations
In view of Sir Issac Newton ideas of rate and change, he put forward the fundamental laws of differential calculus and integral calculus approached as the inverse procedure. The nuts and bolts of advanced Mathematics, ModernDay Physics, and other forms of Engineering are the basis of differentiation. Limits and derivatives fill in as the entry point to limits and derivatives for class 11 CBSE students.
Limits of a Function
A limit is defined as a function that has some value that approaches the input. A limit of a function is represented as:
\[\lim_{n\rightarrow c}\] f(n) = L
Let us consider two functions: p and q. Now, assume “a” is a value such that
\[\lim_{x\rightarrow a}\] p(x) and \[\lim_{x\rightarrow a}\] q(x) exists.

\[\lim_{x\rightarrow a}\] [p(x) + g(x)] = \[\lim_{x\rightarrow a}\] p(x) + \[\lim_{x\rightarrow a}\] g(x)

\[\lim_{x\rightarrow a}\] [p(x) – g(x)] = \[\lim_{x\rightarrow a}\] p(x) + \[\lim_{x\rightarrow a}\] g(x)

For every real number k,
\[\lim_{x\rightarrow a}\] [kp(x)] = k \[\lim_{x\rightarrow a}\] p(x)

\[\lim_{x\rightarrow a}\] [p(x) q(x)] = \[\lim_{x\rightarrow a}\] p(x) x \[\lim_{x\rightarrow a}\] q(x)

\[\lim_{x\rightarrow a}\] \[\frac{p(x)}{q(x)}\] = \[\frac{\lim_{x\rightarrow a} p(x)}{\lim_{x\rightarrow a} q(x)}\]
To learn more about limits, check out the maths limits formulas given below.
Limit Formulas

\[\lim_{x\rightarrow 0}\] \[\frac{sinx}{x}\] = \[\lim_{x\rightarrow 0}\] \[\frac{tanx}{x}\] = 1

\[\lim_{x\rightarrow 0}\] \[\frac{sin^{1}}{x}\] = \[\lim_{x\rightarrow 0}\] \[\frac{tan^{1}}{x}\] = 1

\[\lim_{x\rightarrow 0}\] \[\frac{ln(1+x)}{x}\] = 1

\[\lim_{x\rightarrow 0}\] \[\frac{a^{x}1}{x}\] = lnₐ

\[\lim_{x\rightarrow 0}\] \[\frac{e^{x}1}{x}\] = 1

\[\lim_{x\rightarrow a}\] \[\frac{x^{n}a^{n}}{(xa)}\] = n.\[a^{n1}\]

\[\lim_{x\rightarrow 0}\] \[(1+x)^{\frac{1}{x}}\] = \[\lim_{x\rightarrow 0}\] \[(1+x)^{\frac{1}{x}}\] = \[\lim_{x\rightarrow \infty}\] \[(1+\frac{1}{x})^{x}\] = e

\[\lim_{x\rightarrow 0}\] \[\frac{(1+x)^{m}1}{x}\] = m
Solves Problems
Here are a few limited solved problems.
Question 1: Find \[\lim_{x\rightarrow 4}\] x+4
Solution:
\[\lim_{x\rightarrow 4}\] x+4
= 4 + 4
= 8
Question 2: Find \[\lim_{x\rightarrow 4}\] 2x+4
Solution:
\[\lim_{x\rightarrow 4}\] 2x+4
= 4(2) + 4
= 8 + 4
= 12
Derivatives
The rate at which a quantity instantaneously changes to another quantity, it is known as a derivative.
Definition of Derivatives Using Limits
A derivative function f at y = a is the limit of the slope. Now, the limit of this slope of this line from y = a to y = a + c as c approaches 0. This symbolises that the limit [f(a) – f(a+c)]/c as c → 0.
Basic Concepts of Derivatives
Below are formulas that are related to inverse, hyperbolic, and trigonometric functions:

\[\frac{d}{dx}\] [ p(x) + q(x)] = \[\frac{d}{dx}\](p(x)) + \[\frac{d}{dx}\](q(x))

\[\frac{d}{dx}\] [ p(x) – q(x)] = \[\frac{d}{dx}\](p(x)) – \[\frac{d}{dx}\](q(x))

\[\frac{d}{dx}\] [ p(x) × q(x)] = \[\frac{d}{dx}\] [(p(x))]q(x) + p(x)\frac{d}{dx}[q(x)]

\[\frac{d}{dx}\] \[[\frac{p(x)}{q(x)}]\] = \[\frac{\frac{d}{dx} [(p(x))]q(x)p(x)\frac{d}{dx}[q(x)]}{(q(x))^{2}}\]
Example: Find the derivative of the sin y at y = 0.
Solution: Consider, f(y) = sin y
then, f'(0) = lim_{y→0} [f(0+h) – f(0)]/y
= lim_{y→0} [sin(0+y) – sin(0)]/y
= lim_{y→0} [sin y]/c
= 1
1. What is the Limit of a Function?
A limit is defined as a function that has some value that approaches the input. A limit of a function is represented as:
lim _{n}_{→c }f(n) = L
2. What are Derivatives?
The rate at which a quantity instantaneously changes to another quantity, it is known as a derivative. A derivative function f at y = a is the limit of the slope. Now, the limit of this slope of this line from y = a to y = a + c as c approaches 0. This symbolises that the limit [f(a) – f(a+c)]/c as c → 0.
3. What are Formulas Related to Other Trigonometric Functions?

d/dx [p(x) + q(x)] = d/dx(p(x)) + d/dx(q(x))

d/dx [p(x) – q(x)] = d/dx(p(x)) – d/dx(q(x))

d/dx [p(x) x q(x)] = d/dx [p(x)]q(x) + p(x) d/dx [q(x)]

d/dx [p(x)/q(x)] = {d/dx[p(x)]q(x) – p(x) d/dx [q(x)]} / (g(x))^{2}