Chapter 2 - Limits,Continuity and Derivatives 0 PDF

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Notes on limits, continuity and derivatives...

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Chapter 2- Limits, Continuity and Derivatives 2.1 Limits Let f  x  be a function of x . To obtain the value of f  x  at any point say x a , we substitute the value a for x in f  x  . Example 2.1.1 f  x  2x ,

for x 3, f  3 2 3 6 .

Example 2.1.2 x Suppose f  x  

2

 2 x 1 12  2 1 1 0  , which is undefined. . Then if x 1 , f  1  x1 1 1 0

What do you do in this case? We may circumvent this problem by finding out what happens to f  x  as x gets closer and closer to the value 1 . x Consider the table for values of f  x  

f  x

x

-1

0 0.1 0.2

 0.9  0.8  0.7

0.3 0.4 0.5

 0.6  0.5

2

 2 x 1 , x1

x

f  x

x

0.6 0.7 0.8

 0.4

2 1.9 1.8

1 0.9 0.8

1.7 1.6 1.5

0.7 0.6 0.5

0.9 0.99 0.999

 0.3  0.2  0.1  0.01  0.001

f  x

f  x

x 1.4 1.3 1.2

0.4 0.3 0.2

1.1 1.01 1.001

0.1 0.01 0.001

So as x approaches 1 either from the right or from the left of 1 f  x  approaches 0 . We then say that f  x  has a limit of 0 as x approaches1 . Example 2.1.3 1 x

Suppose g  x   , for x 0 . x

g  x

x

g  x  2.5

1  0.9  0.8  0.7  0.6

-1

 0.4

 1.1111  1.25  1.4286  1.6667

 0.3  0.2  0.1  0.01

 3.3333 5  10  100

 0.5

2

 0.001

 1000

x

g  x

x

g  x 2.5

1 0.9 0.8 0.7

1 1.1111 1.25 1.4286

0.4 0.3 0.2 0.1

3.3333 5 10

0.6 0.5

1.6667 2

0.01 0.001

100 1000

8

Here g  x  does not appear to get close to any value as conclude that g  x  has no limit as x approaches 0 .

x

approaches 0 . We therefore

Definition In general we say that the limit of f  x  as x tends to c is L if f  x  gets close to L f  x   L . In Example 2.3 we find that when x is close to c . This can be written as lim x c

as x approaches 0 from the negative (left) side, g  x  gets close to minus infinity, while as x approaches 0 from the positive (right) side g  x  becomes very large. This g  x    and lim g  x   . Thus, there is no limit of g  x  as is to say that xlim  0 x 0 approaches 0 .

x

Example 2.1.4 x  2 7  x

Suppose f  x  

for x  3 for x 3

f  x  4 and lim f  x  1 . Hence lim f  x does not exist. Example We see that xlim  x 3 3 x 3

2.4 is an example of a one-sided limit. The limit exists if both one-sided limits exist and are equal. The graph of the above function is as follows: x

f  x

x

f  x

2 2.4 2.6 2.8 2.9

0

4 3.6 3.4 3.2

3

2.99

0.99

0.4 0.6 0.8 0.9

3.1

3.4 3.6 3.8 3.9

3.0

4

f x

x

9

This function does not have a limit at a point 3. It consists of two straight lines that will never touch each other. There is a jump at x 3 , and we say that the function f  x  is discontinuous at x 3 . Its value at x 3 may be defined as 1 or 4 or any value between 1, 4  . 2.1.1 Rules for computing limits (i)

k k . For example, let f  x  5 . Obtain the If k is a constant function, then lim x c 5 5 . limit of f  x  as x  2. That is lim x 2

(ii)

f  x   L, then lim kf  x  k lim f  x  k L . For example If lim x c x c x c lim 3  x  2  3 lim  x  2  3  4  2  3  6  18 . x 4

(iii)

x 4

f  x   L and lim g  x  M , then If lim x c xc lim  f  x  g  x   lim f  x  lim g  x  L M x c

(iv)

x c

x c

f  x f  x  xlim L  c  , provided M 0 x c g  x  lim g  x  M lim

x c

(v)

lim  f  x  g  x    lim f  x  lim g  x   L M x c

x c

x c

P

(vi)

P P lim  f  x     lim f  x   L , where P is constant.  x c  x c

Example 2.1.5 a)

 x  5 6 x 5 lim  x 1   3 x 1 x  3 lim  x  3  2

lim

x 1









b)

lim x2  1  x  3   lim x2  1 lim  x  3   4  1   2  3  3  5  15 .

c)

lim x2  x  1

d)

 1 lim  x 3   lim x  x 27  x 27  

e)

lim 2 x 20 1 . x 0

f)

lim

x 2

x 4

x 2





2



x 0



x 2

 lim x2  lim  x  1  x 4



1 3

x 4



2

 16  4  1   19  361 . 2

1

 27 3 3

1 1 1 1 1 1    , also lim   0 x  x x lim x 0 lim x  x 0

x 

10

2



g)

4

17 

  x

h)

i)

j)

17   lim 4

 4x  17 lim  x  x   x   4  0 lim    x   45  2 x  x   45   45  0  2  lim 2 2

2 x lim x

x 

2

 3 x 30

2

 4x  3



 

x 

  x

 

4   2 . 2



3 30    x x2  2  0 0  2 . 4 3  10 0  lim  1   2  x   x x 

  lim  2  x 

0 x2  x  6 . Notice that here direct substitution will give us . So 0 x 2 x 2 2 x x  6  x  3  x  2   lim x  3 2  3 5 . factorization is required. So lim  lim   x 2 x2 x2 x2 x2 lim

 x 4  x  1 lim x2  3 x  4  lim   x  1  4  1  5 .     x x x  4 4 4 x 4 x 4 lim

2.2 Continuity If we can draw the graph of y  f  x for different values of x without lifting the pen, then, f is said to be a continuous function. f  x   f  c  . That is, Definition: f  x  is said to be continuous at a point x c if lim xc lim f  x  lim f  x  f  c . In this case lim f  x  f  c , for all c . x c x c

x c

Example 2.2.1 The following represent continuous functions.

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Example 2.2.2 f  x

is discontinuous at x 2 .

Example 2.2.3

f  x f  x

is discontinuous at x c and continuous in all other points.

2.2.1 Rules for continuous functions Let f  x  and g  x  be continuous at x c , and let k be a constant. Then, (1) the sum of the continuous function is continuous, that is, f  x   g  x  is continuous at x c . (2)

the difference of continuous function is continuous; f  x   g  x  is continuous at x c .

(3)

the product of continuous function is continuous; f  x  g  x  is continuous at x c . the ratio of continuous function is continuous as long as the denominator is not

(4)

zero;

f  x

g  x

is continuous at x c , if g c  0 .

(5)

the function f  x   xn is continuous everywhere for all possible integer values of n .

(6)

the constant function f  x  k is continuous everywhere.

(7)

all polynomials are continuous.

Example 2.2.3

12

Discuss the continuity of the following functions. (a) f  x 4 x  3

(b)

g  x  4 x3  5 x2 12 x  3

(c) h  x   5 x  3  3x 2  3 x  9  . Solution 2.2.3 (a)

Since f  x 4 x  3 is a linear function, it is a polynomial of degree 1 and therefore continuous everywhere.

(b) It is a polynomial of degree 3 and therefore continuous everywhere. (b)

The function is the product of polynomials, and so it is the product of two continuous functions. Therefore it is continuous everywhere.

Example 2.2.4 Discuss the continuity of the following functions: (a)

f  x 

x 3 2 x

(b)

u x 

x2  25 . x2  3 x 10

Solution 2.2.4 (a)

The function is the ratio of two polynomials, and since they are continuous everywhere, f  x  is continuous everywhere except where the denominator is zero, this occurs only at x 2 . Thus f  x  is continuous everywhere except at x 2 . u x 

(b) x  1

2 x  25 is continuous everywhere except at x  3 x 10 2

and x 5 .

Note: Be careful when discussing the continuity of rational functions. Do not factor the numerator and denominator and then cancel. Remember that the function is discontinuous at every point where the denominator is zero.

2.3. Differentiation The central concept in the study of Calculus is the concept of a derivative. The process of finding the derivative of a function is called differentiation. Definition: The derivative of a function y  f  x  , with respect to x , is defined by f  x h   f  x  dy d assuming that the limit exists. This is   f  x    f   x  lim h 0 dx dx h termed as differentiation from first principles.

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Example 2.3.1 Find the derivative of f  x  4 x  2 . Solution 2.3.1 f  x h  f x  4  x  h    4x  2  lim f   x  lim h 0 h 0 h h 4 x 4 h 2  4 x  2 4h lim lim lim 4 4 . h 0 h 0 h h 0 h Example 2.3.2 Differentiate the following function from first principles (use the definition): f  x  3 x  2 x 2 . Solution 2.3.2 f  x  h  3  x  h   2  x  h  3 x  3h  2  x2  2 xh  h2  2

3 x  3h  2 x2  4 xh  2 h2 . f  x  h   f  x  3x  3h  4 xh  2h2   3 x  2 x2  3 x  3h  2 x2  4 xh  2 h2  3 x  2 x2 3h  4 xh  2 h2 h  3  4 x  2h  .

So f   x  lim h 0

f  x  h  f  x  h  3  4 x  2h  lim lim 3  4 x  2h  3  4 x . h 0 h 0 h h

Example 2.3.3 Find f   x  from first principles (use the definition) in each of the following: 3 (i) f  x  3 x  2 x 1 , (ii) f  x  x 2  2 , (iii) f  x  2 x and (iv)

f  x 12 x  2 x 3 .

Solution 2.3.3

' Definition: If f  x  does not exist at a point c , then f  x  is said to be not differentiable.

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Example 2.3.4 Show that f  x   x is not differentiable at x 0 . Solution 2.3.4 f  xh  f  x f 0  h   f 0  h 0 h . lim lim lim f   x  lim    h 0 h h h 0 0 0 h h h h h 1 if h  0, h h  But h  h  1 if h  0.  h h h Therefore lim 1 and lim  1. Hence f  x   x is not differentiable at x 0 . h 0 h h 0 h

2.4 Tangent Lines and Secant Lines 2.4.1 Tangent Lines The derivative of a function f at a value x0 is the slope of the line tangent on the graph of f at the point  x0 , f  x0   . The tangent line to the curve y  f  x  at the point P  x0 , f  x0   is the line through f  x  f  x0  , provided that this limit exists. P with slopem  f  x 0   lim x x0 x  x0 Example 2.4.1 Find the equation of the tangent line to the parabola y  x 2 at the point  1,1 . Solution 2.4.1 f  xh  f x  x  h   x 2 lim x 2  2xh h 2  x 2 lim h  2x  h  2 x 2 lim h 0 h 0 h 0 h 0 h h h h at x 1 . So the equation of the tangent line at  1,1 is y  1 2  x  1 or y 2 x  1 . 2

m  lim

Example 2.4.2 Find the equation of the tangent line to the hyperbola y  Solution 2.4.2

15

3 at the point  3,1  . x

2.4.2 Secant Lines A straight line joining any two distinct points on a curve y  f ( x) is called a secant line. The slope of the secant line passing through the points ( x0  h, f ( x0  h) ) and ( x0 , f ( x0 )) is given by f ( x 0  h)  f ( x 0) m . (x 0  h )  x 0 The equation to the secant line passing through the points ( x0  h, f ( x0  h) ) and ( x0 , f ( x0 )) is given by y  f ( x0 ) m( x  x0 ).

Example 2.4.3 Find the slope and the equation of the secant line passing through the points (1, 3) and (3, 5). Example 2.4.4 Find the slope and the equation of the secant line on a curve y 2 x 2  3 x  3 at x  1 and x 2.

2.5 The Derivative as a Rate of Change Consider the line given by the equation y mx  c : m slope , c y  int ercept . y Slope is defined as the change in y divided by the change in x. That is m  . For x m is the same along the line. Hence the slope of a straight the line above, the ratio line could be referred to as the rate of change of y w.r.to. x . Given a function f , we associate with it a new function f  , called the derivative of f , defined by f  x h   f  x  f  x  lim . h 0 h The derivative f   x 0  measures the rate of change of f  x  at x  x0 . If f  x  denotes the distance traveled by satellite at time t , then f  t  is its velocity and f   t  is its acceleration.

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