THE DERIVATIVE I by suhalia

MODULE 1: THE DERIVATIVE

DEFINITION OF LIMIT, COMPUTING LIMITS

• The exact value of f when x = a is f(a) . • Limit – The behavior of f(x) as x closer and closer to a value of a. •

the notation;lim f(x) x a

L

A function f(x) approaches a limit L as x approaches a ; If the value of f(x) can be made as close to L as we please by taking x sufficiently near, though not equal to a.

Example 1 Complete the following table. Discuss the behavior of the values of f(x) when x is closed to 1. What are the value of f(x) when x approaches 1 from the left i.e. through values that are less than 1 (x<1)? Ans:3 x approaches 1 from the right i.e. through values that are more than (x>1)? Ans:3 x f(x) = x2 + 2

0.9

2.81

0.999 0.9999 1 1.0001 2.9801 2.998001 2.99980001 3 3.00020001 0.99

x approaching 1 from lef

1.001 1.01 3.002001 3.0201

1.1 3.21

x approaching 1 from right

lim x 2 ď&#x20AC;Ť 2 x ď&#x201A;Ž1

The graph of f ( x)  x 2  2 y= f(x)

y = x2 + 2



x 1



x 1

lim ( x 2  2)  lim ( x 2  2)  3

x 1

Example 2 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 x3  1 f ( x)  x 1

0.5

0.75

1.75

2.3125

1.001 1.01 3.003001 3.0301

0.9

0.99

0.999

2.71 2.9701 2.997001 1.1 3.31

1.25 3.8125

1 ? 1.5 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 x approaches 1 from the right i.e. through values that are more than 1? Ans: 3

x3  1 f  x  x 1

y 8 6 4

3 2 x

x approaches 1 from the left

x approaches 1 from the right

The value of f(x) approaches 3 when x approaches 1 from the left and f(x) approaches 3 when x approaches 1 from the right.

Definition of Limits â&#x20AC;˘

Intuitive Meaning: Graphical Interpretation We can make f(x) as close to L y=f(x) as desired by choosing x sufficiently close to a, and xď&#x201A;ša. L

Note: For the limit to exist, the values of f(x) must approach the same number on both sides.

One Sided Limits

ii. lef - hand limit

lim- f  x   L

i.right - hand limit

lim f  x   L

x a

iii Two sided Limits 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. lim f(x)  L if and only if x a

i. lim f(x) exist x a

ii. lim- f(x) exist x a

iii. lim f(x)  lim- f(x)  L x a

x a

*In order for a two-sided limit to exist, the limit from the right and the limit from the left need to be the same.

Definition of Limits Example 3 • For the functions in the figures below, find the one sided and two sided limits at x=1 if it exist

lim f ( x )  1 

x 1

lim f ( x )  lim f ( x )

lim f ( x )  3 

lim f ( x )  lim f ( x )

x 1

lim f (x )  3 

x 1

 lim f ( x ) Does not exist  lim f ( x ) Does not exist x1 x1

f (1)  2

f (1)  1

lim f ( x )  3 

x 1

 lim f ( x) DNE x1

f (1) DNE

Definition of Limits Example 4 • For the functions in the figures below, find the one sided and two sided limits at x=2 if it exist

lim f ( x )  2

x2

lim f ( x )  2

x 2

lim f ( x )  lim f ( x )  2

x 2 

x 2 

 lim f ( x )  2 x2

f (2)  DNE

lim f ( x )  2

x 2 

lim f ( x )  2

x2

lim f ( x )  lim f ( x )  2

x  2

x  2

 lim f ( x )  2 x 2

f ( 2)  3

lim f ( x )  2

x 2 

lim f ( x )  2

x 2

lim f ( x )  lim f ( x )  2

x  2

x 2

 lim f ( x )  2 x2

f ( 2)  2

Definition of Limits

Example 5 (Practice) • For the functions in the figures below, find the one sided and two sided limits at (a) x= -1, and (b) x=0 if it exist. y 3

-2

(a) lim f ( x )  0 x  1

-1

x f ( x )  1 (b) xlim 0  lim f ( x )  0

lim f ( x )  0

x 0 

lim f ( x )  lim f ( x )  0

x 0 

x  1 x  1

x  1

 lim f ( x )  0 x  1

f ( 1)  0

lim f ( x )  lim f ( x ) x 0

 lim f ( x ) DNE x0

f ( 0)  0

Definition of Limits Example 6 • For the function graphed in the accompany figure, find lim f ( x ) (ii) lim f ( x ) (iii) lim f ( x ) (i) x 0  x 2  x 2

lim f ( x ) (iv) x 2

(v) f(2)

(i ) lim f ( x )  7

x 0

lim f ( x ) DNE

x 0 

lim f ( x ) DNE x 0

(ii ) lim f ( x )  1 x2

(iii ) lim f ( x )  1

1 2

x 2

(iv ) lim f ( x )  1 x 2

( v ) f ( 2)  4

Properties of Limits Assume that k is a constant Constant rule: lim

x a

Limit of x rule: Multiple rule : Sum and Difference rule:

k k

lim x  a

x a

lim k f ( x )  k lim f ( x )

x a

lim f ( x )  g( x )  lim f ( x )  lim g( x )

x a

Properties of Limits Continue; Product rule : Quotient rule:

Power rule:

lim

x a

 f ( x ) g( x )   lim f ( x ) x a

lim f ( x ) f ( x ) x a lim g( x )  x a lim g( x ) x a

  g ( x )  xlim    a 

if lim g(x)  0 xa

lim n f(x)  n lim f(x) x a

x a

if lim f(x)  0 when n is even x a

Example 7 Evaluate the following limits: lim 15 (b) lim x (c) lim 3x  5 (a) x 2 x 1 x  3

( 2x  5) 4 (f) lim (e) xlim x  2 2

2x  1 3x  4

Solution (a)

lim

15  15

lim x a

k k

lim

x  3

lim x a

xa

lim

3x  5  lim 3 x  lim 5

x 2

(b)

x  3

(c)

x 1

 3 lim x  lim 5 x 1

 3 1  5  2

x 1

(d)

(g) lim

x 3

lim (2x  3)( x  4)

x 3

x 2  5x  2

Example 7 (Solution)     ( 2 x  3 ) ( x  4 ) ( 2 x  3 )( x  4 ) lim lim lim  x3   x3  (d) x 3     lim 2 x  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   x 3

  2 3  3  3  4

 9 4

(e)

lim ( 2x  5)

x 2

   lim  2 x  lim 5 x 3   x 2 4      2 lim x  lim 5 x 3   x 2

   2 2   5 1

EVALUATION OF LIMITS

Techniques for evaluating limits

Substitution Factorization Rationalization

Polynomials

p( x) q( x) Rational Function

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

Note: For values of the function for which f(a) is defined, the limit can be found by substitution Example 8 3 x  5 x  1 (b) lim Evaluate (a) lim x 2

x  2

3x 2  6

B. Limits of Rational Functions Theorem: Let

p( x ) q( x )

f(x) =

be a rational function and a be any real number

i. If q(a) 0, ii. If q(a)=0 but

lim f ( x )  f (a)

x 0, a p(a)

then

does not exist

lim f ( x )

xa

Example 9 Find the following limits: (a)

3x 2  2 lim x 2 x 1

2 5 x  x 1 (b) lim x 3 x3

B. Limits of Rational Functions a. Limit does not exist Note: When the denominator [q(a)=0] is zero, but the limit of the numerator [ p(a) 0] is not zero, the response does not exist can be elaborated upon one of the three ways: (i) The limit may be + (increases without bound) (ii) The limit may be - (decreases without bound) (iii) The limit may be + from one side and - from the other side ( the two sided limit does not exist)

B. Limits of Rational Functions a. Limit does not exist

1   lim x  a x a 1   limx a x  a 

lim x a

1 DNE xa

1 lim ( x  a)2   x a1 lim ( x  a)2   x  a 1  lim   2 ( x  a ) x a

lim f ( x )   and/or lim f ( x )   , then x  a

x a 

x a 

1 lim - - ( x  a)2   x a 1 lim - ( x  a)2   x  a 1  lim   2 ( x  a ) x a

is a vertical asymptote of f(x).

B. Limits of Rational Functions a. Limit does not exist Example 10 (a)Find the following limits: i.

lim

x 3

1 x3

ii.

lim

x 3

1 ( x  3) 2

iii.

lim - (

1 2 ) x3

lim - (

1 2 ) x4

x 3

Find the following limits: lim

iv. x 3

x3 x2  9

lim

x 4

3x ( x  4)

vi.

x 4

B. Limits of Rational Functions b. Indeterminate Limit of the form â&#x20AC;&#x153;zero over zeroâ&#x20AC;? â&#x20AC;˘ If the limits for both the numerator and denominator are zero at point a, i.e. q(a) =p(a)=0 then (i) factorization techniques is used. We factorize and cancel common factors in the numerator and denominator which reduce the fraction to one whose denominator is no longer zero. (ii) rationalizing an expression is used when a radical appears in the function. (multiplying with a suitable conjugate (conjugate surds -refer page 12 Matriculation Mathematics I)

Note : 0

is called an indeterminate form

b. Indeterminate Limit of the form “zero over zero” Example 11 (Factorising) Find the following limits: a. b. x2  9 lim x 3 x  3

2 c. d.x  6 x  7

lim

x 7

Example 12 (Rationalizing) Find : a. b. c. lim

x 3

x 3 x3

lim

x7

lim

x 0

x 4

x x  16  4

2x  8 x 2  x  12

lim

x 1

x 2  3 x  10 lim 2 x  5 x  10 x  25

x8 3 3( x  1)

c. Limits of Piecewise-Defined Functions Piecewise function is a function f(x) defined piecewise, that is f(x) is given by different expressions on various intervals Example 13 x2-1 Let f(x) = x-2

;

x  -1

;

-1 < x  2

(1/2) x Find (a) lim (b) f (x) x  1

;

x>2

lim f ( x )

x2

Limit at Infinity Definition • Recall that the limit of f(x) as x approaches some value is simply the y-value that the function approaches. • A limit at infinity is the value of a function f(x) is approaching as the independent variable, x gets infinitely large in either the positive or negative direction. • 2 cases: i. The function f(x), approaches a numerical value. ii. The function f(x),approaches infinity in either direction.

A. Polynomial Functions Theorem: Let k be a real number. lim k  k , lim k  k a) x  x  

lim x  

x  

lim x  

x 

lim x n  

x 

lim x n  

x  

lim x n  

if the power, n is even if the power, n is odd

x  

• The end behaviour of a polynomial matches the end behaviour of its highest degree term, i.e. if cn ≠ 0 then

lim (c 0  c 1x  ...  c n x n )  lim c n x n

x 

x  

• If lim f ( x )  L , then y  L is a horizontal asymptote of f(x). x 

Example 14 Evaluate the following limits: 5 3 lim 7 x  4 x  2x  9 a) x  

(n  5 is odd)

b) lim 4 x 8  17 x 3  5 x  1 x  

(n  8 is even)

B. Rational Functions i.

lim

Limit of Rational Function Using

 Consider the function f(x) =

for

x 

1 0 x

1 . x  x x above table10 1000 the value 10000 The shows that100 when x increases, of f(x) 0.1 0.01 0.001 0.0001 lim

x 



1 x

lim

f(x)100000 approaches zero i.e.

0.00001

1 0 x

1 Consider the function f(x) = 1 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 decreases, the value of 1 0 f(x) approaches zero i.e. lim x   x

Note: To evaluate limits at infinity, divide the numerator and the denominator by : the highest power of x that occurs in the denominator. Example 15 Find the following limits: a) b)

lim

c) x 

lim

d)x   

lim

x  

4x  3 1 x 4x 3  3 x 2  1 x ( x 2  2) 4x 2  x 2x 3  5

2x 2  3 lim x  1  x

CONTINUITY

Note:A function f is said to be continuous at x = c provided the following conditions are satisfied i. lim Exists f(x) 2-sided limit xc

ii. f(c)

is defined

Exact value

iii. lim f(x)  f(c) x c

If one or more conditions fails to hold, f has a discontinuity at x= c.

Note: When one or more conditions of this definition fail to hold, then we said that f has a discontinuity at x=c. The figures below illustrate a discontinuity at x = c L

lim f ( x )  lim f ( x )  L

x c 

x c 

i.Thus lim f ( x )  L x c

lim f ( x )  lim f ( x )

x c 

x c 

i. lim f ( x ) does not exist xc

ii.f(c) is not defined f(x) is not continuous at x=a.

lim f ( x )  lim f ( x )  a

x c 

x c 

i. lim f ( x )  a x c

ii.f(c)  b, f(c) is defined  f(x) is not continuous at x=a.

iii. lim f ( x )  f ( c ) x c

 f(x) is not continuous at x=a.

Properties of Continuous Functions If the functions f and g are continuous at x=c, then (a) (b) (c) (d)

[f (x)]n is continuous at x=c, f ± g is continuous at x=c, fg is continuous at x=c, f /g is continuous at x=c, if g(c)  0 and has discontinuity

Theorem: A polynomial is continuous for all x. Example 16 (Continuity of polynomial function) Determine whether the following functions are continuous at x = 3 (a)f(x)=x2 – 4x (b) f(x) = x3 -2

at c if g(c) =0

Note:

If f is a rational function, a power function, or a trigonometric function, then f is continuous at any number x = c for which f(c) is defined.

Theorem : A rational function is continuous for all x except those values that makes the denominator is zero. Example 17 (Continuity of rational function) Determine whether the following functions are continuous at the indicated point. 3 a. f (x )  2 ;x  2 x 1

x2  x  2 ;x  1 b. f ( x )  x 1

Example 18 x2  1 If f (x )  3 2 , find the discontinuities of f. x  x  2x