1. Limits and Derivatives
 lim f ( x) is the expected value of f at x = a, given the values of f near x to the
xa
left of a. This value is called the left hand limit of f(x) at a.
 lim f ( x) is the expected value of f at x = a, given the values of f near x to the
xa
right of a. This value is called the right hand limit of f(x) at a.
 If the right and left hand limits coincide, we call that common value the limit of
f(x) at x = a, and denote it by lim f ( x) .
x a
 x 2  5, if x  2
Example: Evaluate lim f  x  where f  x   
x 2
3  3x, if x  2
Solution:
lim f  x   lim  x 2  5    2   5  9
2
x 2 x 2
lim f  x   lim  3  3 x   3  3   2   9
x 2 x 2
Since lim f  x   lim f  x   9, lim f  x   9.
x 2 x 2 x 2
 For a function f and a real number a, lim f ( x ) and f(a) may not be the same.
x a
 Algebra of limits: Let f and g be two functions such that both lim f ( x) and
x a
lim g ( x) exist. Then,
x a
 lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)
x a x a x a
 lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)
x a x a x a
 lim [( f )( x)]   lim f ( x) , where  is a constant
x a x a
f ( x) lim f ( x)
 lim  x a , where lim g ( x)  0
x a g ( x) lim g ( x) x a
x a
 Some standard limits:
 For a polynomial function f(x), lim f ( x)  f (a).
x a
g ( x)
 If f(x) is a rational function of the form f ( x)  , where g(x) and h(x) are
h( x )
g (a)
polynomial functions, such that h(x)  0, then lim f ( x)  .
xa h(a )

2. xn  an
 lim  na n 1 , where n is a positive integer or any rational number
xa xa
and a is positive
sin x
 lim 1
x 0 x
1  cos x
 lim 0
x 0 x
sin 2 x
Example 1: Evaluate lim
x 0 (1  cos x)sin 3 x
Solution: We have
sin 2 x
lim
x 0 (1  cos x ) sin 3 x
 (1  cos 2 x) x
 lim    ( sin 2 x  1  cos 2 x)
x 0
 (1  cos x) sin 3 x x 
 (1  cos x)(1  cos x) x 
 lim  
x 0
 x(1  cos x) sin 3 x 

1  cos x 1 3x
 lim  lim
x 0x 3 x 0 sin 3 x
1 1  1  cos x 
 0   lim  0
3
lim
sin 3 x  x 0 x 
3 x 0 3x
 Since x  0,3 x  0 
1 1
 0   
3 1   lim sin 3 x  1
 3 x 0 

 3x 
0
3
(4  x) 2  8
Example 2: Evaluate lim
x 0 x
0
Solution: The given limit is of the form .
0
Put 4 + x = y so that y  4 [as x  0]

3. 3 3 3
(4  x)  8
2
y  (4) 2 2
lim  lim
x 0 x y 4 y4
3 3 1  xn  an 
 (2) 2  lim
 x a  na n 1 
2  xa 
3 2

2
4 x 2  36
Example 3: Evaluate lim
x 3 x2  x  6
0
Solution: The given limit is of the form .
0
We have
4 x 2  36 4( x 2  9) 4( x  3)( x  3)
lim  lim 2  lim
x 3 x  x6
2 x 3 x  3 x  2 x  6 x 3 x ( x  3)  2( x  3)
4( x  3)( x  3)
 lim
x 3 ( x  2)( x  3)
4( x  3)
 lim , as x  3
x 3 x2
4  6 24
 
3 2 5
 Let f and g be two real-valued functions with the same domain, such that f(x) 
g(x) for all x in the domain of definition. For some a, if both lim f ( x) and
x a
lim g ( x) exist, then lim f ( x)  lim g ( x) .
x a x a x a
 Sandwich Theorem:
Let f, g, and h be real functions such that f(x)  g(x)  h(x) for all x in the
common domain of definition. For some real number a, if
lim f ( x)  l  lim h( x) , then lim g ( x)  l .
x a x a x a
1
Example: Evaluate lim x3 cos  
x 0 x  
1
Solution: lim cos   does not exist.
x 0 x  
1 1
 lim x3 cos    lim x3  lim cos   is not applicable.
x 0
 x  x 0 x 0
 x

4. 1
It is known that 1  cos    1 .
x  
1
 x3  x3 cos    x3
 x
Now, lim   x3   lim x3  0
x 0 x 0
1
 By sandwich theorem, lim x3 cos    0
x 0 x  
 Suppose f is a real-valued function and a is a point in its domain of definition.
The derivative of f at a [denoted by f′(a)] is defined as
f ( a  h)  f ( a )
f '(a)  lim , provided the limit exists.
h 0 h
Derivative of f(x) at a is denoted by f′(a).
 Suppose f is a real-valued function. The derivative of f {denoted by f ( x ) or
d
[ f ( x)] } is defined as
dx
d f ( x  h)  f ( x )
[ f ( x)]  f ( x)  lim , provided the limit exists.
dx h 0 h
This definition of derivative is called the first principle of derivative.
Example 4: Find the derivative of f(x) = x2 + 2x using first principle of
derivative.
f ( x  h)  f ( x )
Solution: We know that f′(x) = lim
h 0 h
( x  h) 2  2( x  h)  ( x 2  2 x)
 f '( x)  lim
h 0 h
x 2  h 2  2 xh  2 x  2h  x 2  2 x
 lim
h 0 h
h 2  2hx  2h
 lim
h 0 h
 lim(h  2 x  2)
h 0
= 0 + 2x + 2 = 2x + 2
 f′(x) = 2x + 2
 Algebra of derivatives of functions: For the functions u and v (provided u and v
are defined in a common domain),
 (u  v) '  u '  v '
 (uv) '  u ' v  uv '  Product rule 

5.  u  u ' v  uv '
     Quotient rule 
v v2
 Some standard derivatives:
d n
 ( x )  nx n 1 , for any positive integer n
dx
d
 (an x n  an 1 x n 1  ...  a1 x  a0 )  nan x n 1  (n  1)an 1 x n 1  ....  a1
dx
d
 (sin x)  cos x
dx
d
 (cos x)   sin x
dx
d
 (tan x)  sec 2 x
dx
Example 5: Find the derivative of the function f ( x)  (3x2  4x  1)  tan x
Solution:
df ( x) d
 (3 x 2  4 x  1) tan x 
dx dx  
d  d 
  (3 x 2  4 x  1)   tan x   (tan x)   (3 x 2  4 x  1)
 dx   dx 
 d d d 
 3 ( x 2 )  4 ( x)  (1)   tan x  sec 2 x  (3 x 2  4 x  1)
 dx dx dx 
= (6x + 4) tan x + sec2x (3x2 + 4x+ 1)
= 3x2 sec2 x + 2x (2sec2x + 3tanx) + (sec2 x + 4tan x)
x( x 2  1)
Example 6: Find the derivative of
2x  3
Solution:
d d
 x( x 2  1)   (2 x  3)  x( x 2  1). (2 x  3)
d  x( x 2  1)  dx  
 
dx  Using quotient rule
dx  2 x  3  (2 x  3) 2
 x(2 x)  ( x 2  1)  (2 x  3)  x( x 2  1)(2)
 
 Using product rule
(2 x  3) 2
(2 x 2  x 2  1)(2 x  3)  2 x( x 2  1)

(2 x  3) 2

6. (3 x 2  1)(2 x  3)  2 x( x 2  1)

(2 x  3) 2
6 x3  9 x 2  2 x  3  2 x3  2 x

(2 x  3) 2
4 x3  9 x 2  3

(2 x  3) 2