CONTINUITY
FUNCTION
BASIC RULES
LIMIT
LIMIT AS FIRST PRINCIPLES
DERIVATIVE
COMPUTING LIMIT
DEFINITION
IMPLICIT DIFFERENTIATION
RULES OF DIFFERENTIATION
TECHNIQUES OF DIFFERENTIATION
LIMIT OF FORM
CHAIN RULE FACTORIZATION
ADDITION
ALGEBRAIC FUNCTION
SUBTRACTION
EXPONENTIAL FUNCTION
MULTIPLICATION
LOGARITHMIC FUNCTION
QUOTIENT
TRIGONOMETRIC FUNCTION
RATIONALIZATION
Obtain the limit of a function intuitively and computing limits including end behaviour
Determine the continuity of a function based on the definition Find the derivative of a function using the first principle method Find the derivatives using rules of differentiation for polynomial, trigonometric, exponential and logarithmic functions
Find the derivative of implicit function
Limits are used to described how a function behaves as the independent variable x approaches a given value. This concept helps us to describe, in a precise way:
the behaviour of f(x) when x is close to but not equal to a particular value a.
Complete the following table. Discuss the behaviour of the values of f(x)
when x is closed to 1. x
f(x) = x2 + 2
0.9
0.99
0.999
0.9999
1
1.0001
1.001
1.01
1.1
Discuss the behaviour of the values of f(x) when x is closed to 1. x approaching 1 from right
x approaching 1 from left x
0.9
0.99
0.999
f(x) = x2 + 2
2.81
2.9801 2.998001
0.9999
1
1.0001
1.001
1.01
1.1
2.99980001
3
3.00020001
3.002001
3.0201
3.21
What are the values of f(x) when x approaches 1 from the left i.e. through values that are less than 1?
Ans: 3
What are the values of f(x) when x approaches 1 from the right i.e. through values that are more than 1?
Ans: 3
f ( x) x 2 2
lim ( x 2) lim ( x 2) 3 2
2
x 1
x 1
lim( x 2) 3 2
x 1
Evaluate the values of f(x) for the given values of x. Discuss the behaviour of the values of f(x) when x is closed to 1. x
0.5
0.75
0.9
0.99
0.999
1
x3 1 f ( x) x 1
1.001
1.01
1.1
1.25
1.5
x
0.5
0.75
0.9
0.99
0.999
1
x3 1 f ( x) x 1
1.75
2.3125
2.71
2.9701
2.997001
?
1.001
1.01
1.1
1.25
1.5
3.003001
3.0301
3.31
3.8125
4.75
What are the values of f(x) when x approaches 1 from the left
i.e. through values that are less than 1? Ans: 3 What are the values of f(x) when x approaches 1 from the
right i.e. through values that are more than 1? Ans: 3
x3 1 lim 3 x 1 x 1
1 x approaching 1 from left
x approaching 1 from right
Graphical Interpretation y = f(x) A function f(x) approaches a limit L as x approaches a
L
lim f x L
x a
a
Intuitive Meaning We can make f(x) as close to L as desired by choosing x sufficiently close to a, and x a.
lim f x L
x a
WHEN DOES A LIMIT EXIST ? the values of f(x) must approach the same
1
number on both sides.
The existence of a limit at ‘a’ has nothing to
2
do with the value of the function at ‘a’. In fact , ‘a’ may not even be the domain of f.
3
However, the function must be defined on both sides of ‘a’.
Intuitive Meaning
(righthand limit)
We can make f(x) as close to L as desired by choosing x sufficiently close to a, and x>a
Graphical Interpretation Let L be any real number. Suppose that f(x) is defined near a for x > a and that as x gets close to a, f(x) gets close to L, written as
lim f x L
x a
y = f(x)
L lim f x L
x a
a
Intuitive Meaning
(left- hand limit)
We can make f(x) as close to L asdesired by choosing x sufficiently close to a, and x<a
Graphical Interpretation Let L be any real number. Suppose that f(x) is defined near a for x<a and that as x gets close to a, f(x) gets close to L, written as
lim f x L
x a
y = f(x)
L
a
lim- f x L
x a
A function f (x) has a limit as x approaches a if and only if it has left-hand and right-hand limits exists and they are equal.
For the functions in the figures below, find the one sided and two sided limits at x=1 if it exist
Solution f(1)=2
lim f(x) 3
x 1
lim f(x) 1
x 1-
lim f(x) lim f ( x )
x 1-
x 1
Therefore, the limit does not exist at x = 1.
Solution lim f(x) 3
f(1)=1
x 1
lim f(x) 1
x 1-
lim f(x) lim f ( x )
x 1-
x 1
Therefore, the limit does not exist at x = 1.
Solution
f(1)=does not exist
lim f(x) 3
x 1
lim f(x) 1
x 1-
lim f(x) lim f ( x )
x 1-
x 1
Therefore, the limit does not exist at x = 1.
For the function in the figure below, find the one sided and two sided limits at (a) x= -1, and (b) x=0 if it exist.
Solution f(-1)=0
lim f(x) 0
x 1
lim - f(x) 0
x -1
lim f(x) lim f ( x ) 0 x -1
x -1-
lim f(x) 0 x -1
Solution f(0)=0
lim f(x) 0
x 0
lim - f(x) -1
x -0
lim f(x) lim f ( x )
x 0
-
x 0
lim f(x) does not exist x0
For the function graphed in the accompany figure, find (i) lim f ( x )
(ii) lim f ( x )
(iii) lim f ( x )
(iv) lim f ( x )
x0
x 2
y
x 2
x2
4
(v) f(2) 1
Solution (i)
lim f ( x )
y
x0
does not exist
(v) f(2) = 4 4
1 2 (iv) lim f ( x ) 1 x 2
(iii) lim f ( x ) 1 x 2
(ii) lim f ( x ) 1 x 2
PROPERTIES OF LIMITS Assume that k is a constant Constant rule Limit of x rule Multiple rule
Sum and Difference rule
lim k k
x a
lim x a
x a
lim k f ( x ) k lim f ( x )
x a
x a
lim f ( x ) g( x ) lim f ( x ) lim g( x )
x a
x a
x a
PROPERTIES OF LIMITS Product rule
lim f ( x ) g( x ) lim f ( x ) lim g( x ) x a x a x a
Quotient rule
lim f ( x ) f ( x ) x a lim g( x ) x a lim g( x ) x a lim
n
x a
f(x) n lim f(x) x a
if lim g(x) 0 xa ,
Power rule
if lim f(x) 0 when n is even x a
Evaluate a) b)
c)
lim k k
lim 15 15
x a
x 2
lim x
x 3
lim x a
3
x a
lim 3 x 5 lim 3 x lim 5 x 1
x 1
x 1
lim k f ( x) k lim f ( x)
x a
3 lim x lim 5 x 1
x 1
= 3(1) – 5 = - 2
x a
lim f ( x ) g( x ) lim f ( x ) lim g( x ) x a x a x a
d)
lim (2x 3)( x 4) lim (2 x 3) lim ( x 4) x 3
x 3
lim f ( x) g( x) lim f ( x) lim g( x)
x a
x a
x a
x 3
( lim 2x lim 3) ( lim x lim 4) x 3
x 3
x 3
x 3
(2 lim x lim 3) ( lim x lim 4) x 3
x 3
x 3
= [2(3) + 3].[3 – 4] =-9
e)
lim x2
(2 x 5) 4
(2(2) 5) =1
4
x 3
f) lim x 2
2x 1 3x 4
lim (2 x 1)
x 2
lim (3x 4)
x 2
lim 2x lim 1
x 2
lim 3 x lim 4
x 2
x 2
x 2
2 lim x lim 1 x 2
x 2
3 lim x lim 4 x 2
2( 2) 1 3( 2) 4 5 2
x 2
lim f ( x ) f ( x ) x a lim g( x ) x a lim g( x ) x a
if lim g(x) 0 xa
How do you evaluate limits?
Evaluate means you should give a numerical answer
A. LIMITS OF POLYNOMIAL FUNCTIONS Theorem: If f is a polynomial function and a is a real number, then
lim f ( x ) f (a)
x a
Evaluate a) lim x 3 5x 1 f ( 2) x 2
23 5( 2) 1
NOTE For values of the function for which f(c) is defined, the limit can be found by substitution.
3 b) xlim 2
3x 6 f ( 2) 2
3( 2) 6 2
6
B. LIMITS OF RATIONAL FUNCTIONS Theorem:
p( x ) Let f x be a rational q( x ) function and a be any real number.
LIMITS OF RATIONAL FUNCTIONS p( x ) f x q( x )
CASE 1 : If q(a) 0 then lim f ( x ) f (a) x a
CASE 2 : If q(a)=0 but p(a) 0 then lim f ( x ) does not exist xa
CASE 3 : If q(a) = 0 and p(a) = 0, (indeterminate form)
3x 2 Find lim x 2 x 1 2
TIPS
2 3x 2 3 ( 2 ) 2 lim lim x 2 x 1 x 2 2 1 14 1 2
14
5x x 1 Find lim x 3 x3 2
TIPS
5x 2 x 1 lim lim x 3 x 3 x3
5(3) 2 3 1 33
49 0
q(3) = 0 but p(3) 49 0
5x x 1 lim (does not exist) x 3 x3 2
The limit may be +
The limit may be - The limit may be ±
CASE 1 : limit may be + (increases without bound)
1 y 2 ( x a)
lim x a
lim x a
1 2 x a 1 lim 2 x a ( x a)
1 2 x a
CASE 2 : The limit may be - (decreases without bound)
1 y ( x a) 2
lim x a
1 2 lim x a x a
1 2 x a
lim x a
1 2 ( x a)
CASE 3 :
The limit may be + from one side and -
from the other side (the two sided limit does not exist)
y
lim x a-
1 xa
1 xa
1 lim xa x a
px f x qx If q(a) = 0 and p(a) = 0,
0 f x 0
indeterminate form
x2 9 Find lim x 3 x 3 TIPS
x 2 9 (3 ) 2 9 0 lim x 3 x 3 33 0 x2 9 ( x 3)( x 3) lim lim x 3 x3 x 3 x 3
lim (x 3) x 3
6
Substitution indeterminate form Factorization
x 2 3 x 10
(5) 2 3(5) 10
0 lim 2 lim 2 x 5 x 10 x 25 x 5 (5) 10(5) 25 0
Find
lim
x 3 x 10 2
lim
x 5
x 2x 5 lim 2 x 10 x 25 x 5 x 5 x 5 x 3 x 10 2
x 5
x2 7 lim x 5 x 5 0
x 10 x 25 2
x2 lim x 5 x5
Hence,
lim
x 5
x2 lim x 5 x5
x 2 3x 10 x 10 x 25 2
does not exist.
lim
x 3
x 3 x3
3 3 0 33 0
indeterminate form
Factorization won’t work
Find
lim
x 3
x 3 x3
lim
x 3
x 3 lim x 3 x 3
x 3 x 3 x 3 x 3 x 3 lim x 3 x 3 x 3 1 lim x 3 x 3 1 3 3 1 2 3
Rationalization (Conjugate)
lim
x 0
Find lim
x 0
x x 16 4
0
lim
x x 16 4
x
x 0 x0 16 4 0 16 4 0
0
0 0 16 4 0
lim
x 0
x x 16 4
x x 16 4 lim x 0 x 16 16 lim x 16 4 x 0
0 16 4 8
x 16 4
x 16 4
C. LIMITS OF PIECEWISE FUNCTIONS
Solution f (x) a) xlim 1 lim f ( x ) lim x 2 1 (1) 2 1
x 1
x 1
0 lim f ( x ) lim x 2 1 2
x 1
x 1
3 lim f ( x ) lim f ( x ) x 1
x 1
lim f x x 1
does not exist.
Solution lim f ( x ) b) x 0 lim f ( x ) lim x 2 0 2
x 0
x 0
2 lim f ( x ) lim x 2 0 2
x 0
x 0
2
lim f ( x ) lim f ( x ) 2 x 0
x 0
lim f x 2. x 0
A) POLYNOMIAL FUNCTIONS Theorem: Let k be a real number.
a)
lim k k
x
x b) xlim
lim x n
x
lim x n
x
lim k k
x
lim x
x
If n is even
lim x n If n is odd x
The end behaviour of a polynomial matches the end behaviour of its highest degree term, i.e. if c n 0 then,
lim c 0 c1x ... c n x n lim c n x n
x
If
lim f x L
x
x
, then y = L is a horizontal
asymptote of f (x)
Evaluate the following limits a)
Highest degree
lim 7x 5 4x 3 2x 9 (n 5 is odd)
x
Highest degree
b)
lim 4 x8 17x3 5x 1 (n 8 is even)
x
B) RATIONAL FUNCTIONS 1 1 Consider the function f x for lim . x x x x f(x)
10 0.1
100 0.01
1000 0.001
10000 0.0001
100000 0.00001
The above table shows that when x increases, the value of f(x) approaches zero. i.e. lim 1 0 x
x
1 1 Consider the function f x for lim . x x x x
-10
-100
-1000
-10000
-100000
f(x)
-0.1
-0.01
-0.001
-0.0001
-0.00001
The above table shows that when x decreases,
1 the value of f(x) approaches zero. i.e. lim 0 x x
1 lim 0 x x
To evaluate limits at infinity, divide the
NOTE
numerator
denominator
by
and the
the highest
power of x that occurs in the denominator.
4x 3 Find lim x 1 x
4x 3 4x 3 lim lim x x 1 x x 1 x x highest power of denominator
4x 3 lim x x x 1 x x x 3 4 x lim x 1 1 x
Divide by the highest power of denominator
1 lim 0 x x
4 lim lim 4 4 x 1 x
Alternatively, A QUICK METHOD for finding limits of rational functions as
x or x - The end behaviour of a rational function matches the end behaviour of the quotient of the highest degree term in the numerator divided by the highest degree term in the denominator.
Solution 4x ď&#x20AC; x 2
lim
x ď&#x201A;Žď&#x20AC;ď&#x201A;Ľ
Find
4x ď&#x20AC; x 2
lim
x ď&#x201A;Žď&#x20AC;ď&#x201A;Ľ
2x ď&#x20AC; 5 3
2x ď&#x20AC; 5 3
= lim
x ď&#x201A;Žď&#x20AC;ď&#x201A;Ľ
2x 3 ď&#x20AC; 5
2 = lim đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ =0
highest degree term in the numerator divided by the highest degree term in the denominator.
4x ď&#x20AC; x 2
1 lim ď&#x20AC;˝ 0 x ď&#x201A;Žď&#x201A;Ľ x
CONTINUITY TEST
CONTINUITY
A function f is said to be continuous at x = c
NOTE
provided the following conditions are satisfied i.
f(c) is defined
ii. lim f(x) exist If one or more conditions fails to hold, f has a discontinuity at x = c.
xc
lim f(x) lim- f(x)
x c
x c
iii. lim f(x) f(c) exist x c
If one or more conditions fails to hold, f has a discontinuity at x = c.
i.
f(c) is defined
ii.
lim f(x) exist
iii.
lim f(x) f(c) exist
x c
f(c) is not defined
c i)
xc
lim f ( x ) lim f ( x )
x c
x c
lim f ( x ) exists x c
ii) f(c) is not defined
f(x) not continuous
i. ii.
f(c) is defined
lim f(x) exist xc
f(x) f(c) exist iii. xlim c
c i) lim f ( x ) lim f ( x ) x c
lim f ( x )
x c
x c
Does not exists
f(x) not continuous
i. ii.
f(c) is defined
lim f(x) exist xc
f(x) f(c) exist iii. xlim c
c
f(c)
i) lim f ( x ) lim f ( x ) x c
x c
lim f ( x ) exists x c
ii) f(c) is defined
iii) lim f ( x ) f (c ) x c
f(x) not continuous
PROPERTIES OF CONTINUOUS FUNCTION If the functions f and g are continuous at x=c,then (a) [f (x)]n is continuous at x = c, (b)
f ± g is continuous at x = c,
(c)
fg is continuous at x = c,
(d)
f /g is continuous at x = c, if g(c) 0 and has discontinuity at c if g(c) =0
Theorem
Determine whether the following
A polynomial is continuous for all x.
functions are continuous at x = 3 a) f(x)=x2 â&#x20AC;&#x201C; 4x b) f(x) = x3 -2
a) f(x)=x2 – 4x
Solution
lim x 4 x 3 4(3) 2
2
x 3
9 12 3 lim x 4 x 3 4(3) 2
lim f(x) exist xc
2
x 3
9 12 3
lim f(x) lim- f(x)
x c
x c
lim x 2 4 x 3 x 3
f (3) 3 4(3) 3 2
lim x 4x f (3) 3 2
x3
f(c) is defined
lim f(x) f(c) exist x c
Since lim x 4x f (3) 3 hence f ( x) x 4x 2
x3
is continuous at x = 3
2
If f is a rational function, a power
NOTE
function, or a trigonometric function, then f is continuous at any number x = c
for which f(c) is defined.
A rational function is continuous for
Theorem
all x except those values that makes the denominator is zero.
Determine whether the following functions are continuous at the indicated point
a)
b)
f x
3 x2 1
; x2
x2 x 2 f x ; x 1 x 1
Solution 3 lim 2 1 x 2 x 1
3 lim 2 1 x 2 x 1
3 lim 2 1 x 1 x 2
f 2 1 Since lim f x 1 f 2 x 2
hence f(x) is continuous at x = 2.
Solution x2 x 2 lim x 1 x 1
0 0
x2 x 2 x 1x 2 lim lim x 1 x 1 x 1 x 1 lim x 2
x2 x 2 x 1x 2 lim lim x 1 x 1 x 1 x 1 lim x 2 x 1
x 1
1 2 3
1 2 3
x x2 lim 3 x 1 x 1 2
(1) 1 2 0 f (1) 1 1 0 2
is undefined
2 x2 x 2 Since lim f (1) hence f(x)= x x 2 x 1 x 1 x 1
Is not continuous at x = 1.
If
f x
Solution
x 1 2
x 3 x 2 2x
, find the discontinuities of f.
x2 1 x2 1 f (x) 3 2 x x 2x xx 2x 1 f is discontinuous when the denominator is zero
x( x 2)( x 1) 0 x 0, x 2, x 1 Hence f(x) is not continuous at x = 0, -2, 1.
x 3 y 3 3 xy 9
b)
TIPS use
Solution
dy dy 3x 3y 3 x y 0 dx dx dy 2 2 dy 3x 3y 3x 3y 0 dx dx dy 2 dy 3y 3x 3x 2 3y dx dx dy 3x 3y 2 3x 2 3y dx dy 3x 2 3y dx 3x 3y 2 2
2
(x 2 y) x y2
dy dv du u v dx dx dx