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Limits and continuity

Dec 31, 2015

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Limits and continuity. Start from an example. If the voltage apply to the resistor can be express by the function Determine the maximum voltage across the resistor. We can calculate the maximum or minimum by use derivative. Contents. Concepts of Limits and Continuity - PowerPoint PPT Presentation
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Page 1: Limits and continuity

1

Limits and continuity

Page 2: Limits and continuity

2

Start from an example If the voltage apply to the resistor can be express

by the functionDetermine the maximum voltage across the resistor.

We can calculate the maximum or minimum by use derivative

296 23 xxxy

dx

dy

Page 3: Limits and continuity

3

Contents

Concepts of Limits and Continuity Derivatives of functions Differentiation rules and Higher

Derivatives Applications

Page 4: Limits and continuity

4

Differential Calculus

Concepts of Limits and Continuity

Page 5: Limits and continuity

5

The idea of limits

Consider a function

The function is well-defined for all real values of x

The following table shows some of the values:

2x)x(f

x 2.9 2.99 2.99 3 3.001 3.01 3.1

F(x) 8.41 8.94 8.994 9 9.006 9.06 9.61

92

33

xlim)x(flim

xx

Page 6: Limits and continuity

6

The idea of limits

Page 7: Limits and continuity

7

Concept of Continuity

E.g. is continuous at x=3?

The following table shows some of the values:

exists asand => f(x) is continuous at x=3!

2x)x(f

x 2.9 2.99 2.99 3 3.001 3.01 3.1

F(x) 8.41 8.94 8.994 9 9.006 9.06 9.61

93

)x(flimx

93

)x(flimx

)x(flimx 3

933

)x(flim)x(flimxx

)(f)x(flimx

393

Page 8: Limits and continuity

8

Differential Calculus

Derivatives of functions

Page 9: Limits and continuity

9

Derivative (導數 )

Given y=f(x), if variable x is given an increment x from x=x0, then y would change to f(x0+x)

y= f(x0+x) – f(x)

yx is the slope of triangular ABC

x

f(x0+x)

f(x0)

x0+xx0

y

x

y

Y=f(x)

A

B

C

Page 10: Limits and continuity

10

Derivative

What happen with yx as x tends to 0? It seems that yx will be close to the slope of the

curve y=f(x) at x0. We defined a new quantity as follows

If the limit exists, we called this new quantity as the derivative of f(x).

The process of finding derivative of f(x) is called differentiation.

x

)x(f)xx(flim

x

ylim

dx

)x(df

dx

dyxx

00

Page 11: Limits and continuity

11

Derivative

y

f(x0+x)

f(x0)

x0+xx0x

x

y

Y=f(x)

A

B

C

Derivative of f(x) at Xo = slope of f(x) at Xo

Page 12: Limits and continuity

12

Differentiation from first principle

Find the derivative of with respect to (w.r.t.) x

xx)x(fy 32

32

32

32

3332

33

0

2

0

222

0

22

0

0

0

x

)xx(limx

x)x()x(xlim

x

)xxxx)x()x(xxlim

x

)xx()xx()xx(lim

x

)x(f)xx(flim

x

ylim

dx

dy

x

x

x

x

x

x

To obtain the derivative of a function by its definition is called differentiation of the function from first principles

Page 13: Limits and continuity

13

Differential Calculus

Differentiation rules and Higher Derivatives

Page 14: Limits and continuity

14

Fundamental formulas for differentiation I

Let f(x) and g(x) be differentiable functions and c be a constant.

0dx

)c(d

dx

)x(dg

dx

)x(df

dx

))x(g)x(f(d

dx

)x(dfc

dx

))x(cf(d

1 nn

nxdx

)x(dfor any real number n

Page 15: Limits and continuity

15

Examples

Differentiate and w.r.t. x

143 2 xx 9685 23 xxx

46

0423

043

143

143

2

2

2

x

)x(dx

)x(d

dx

)x(d

dx

)(d

dx

)x(d

dx

)x(d

dx

dy

xxy

61615

062835

968

55

9685

2

2

23

23

xx

)x()x(

dx

)(d

dx

)x(d

dx

)x(d)(

dx

)x(d

dx

dy

xxxy

Page 16: Limits and continuity

16

Fundamental formulas for differentiation II

Let f(x) and g(x) be differentiable functions

dx

))x(f(d)x(g

dx

))x(g(d)x(f

dx

))x(g)x(f(d

2))x(g(dx)x(dg

)x(fdx)x(df

)x(g

dx

))x(g/)x(f(d

Page 17: Limits and continuity

17

Example 1

Differentiate w.r.t. xx

xy

32

32

2

2

2

32

12

32

332332

32

3232

3232

)x(

)x(

))(x())(x(

)x(dx

)x(d)x(

dx)x(d

)x(

dx

dy

x)x(f 32 x)x(g 32

Page 18: Limits and continuity

18

Example 2

Differentiate w.r.t. x)xx)(xx(y 523142 22

2222424

4452326142

422523223142

142523

523142

142523

523142

23

22

22

22

22

22

22

xxx

)x)(xx()x)(xx(

))x()(xx())x()(xx(

)dx

)(d

dx

dx

dx

dx)(xx()

dx

)(d

dx

dx

dx

dx)(xx(

dx

)xx(d)xx(

dx

)xx(d)xx(

dx

dy

Page 19: Limits and continuity

19

Fundamental formulas for differentiation III

)xsin(dx

)xcos(d )xcos(

dx

)xsin(d

22

1))x(sec(

))x(cos(dx

)xtan(d

22

1))x(csc(

))x(sin(dx

)xcot(d

)xtan()xsec(dx

)xsec(d )xcot()xcsc(

dx

)xcsc(d

!

x

!

x

!

x

!

xxewhere x

54321

5432x

x

edx

de

xdx

)xln(d 1 x)eln(where x

ln(x) is called natural logarithm

Page 20: Limits and continuity

20

Differentiation of composite functions

To differentiate w.r.t. x, we may have problems as we don’t have a formula to do so.

The problem can be simplified by considering composite function:

Let so and 12 xy12 xu uy

12 xy

Page 21: Limits and continuity

21

Chain Rule

Chain Rule states that :given y=g(u), and u=f(x)

So our problem and

dx

du

du

dy

dx

dy

2

122

1

2

122

1

1

)x(x)x(u

dx

du

du

dy

dx

xd

dx

dy 2

1

2

1 u

du

ud

du

dy

x

dx

xd

dx

du2

12

u)u(gy 12 x)x(fu

Page 22: Limits and continuity

22

Example 1

Differentiate w.r.t. x

Simplify y by letting so nowBy chain rule

3))x(cos(y

3uy

dx

du

du

dy

dx

dy

23

3udu

du

du

dy )xsin(

dx

))x(cos(d

dx

du

22 33 ))x)(cos(xsin())xsin((udx

du

du

dy

dx

dy

)xcos(u

Page 23: Limits and continuity

23

Example 2

Differentiate w.r.t. x

Simplify y by letting so nowBy chain rule

2xey

2xu uey

dx

du

du

dy

dx

dy

uu

edu

de

du

dy x

dx

dx

dx

du2

2

2

22 xu xexedx

du

du

dy

dx

dy

Page 24: Limits and continuity

24

Example 3

Differentiate w.r.t. x

Simplify y by letting so nowBy chain rule

)xln(y 32 2

32 2 xu )uln(y

dx

du

du

dy

dx

dy

udu

)uln(d

du

dy 1 x

dx

)x(d

dx

du4

32 2

32

44

12

x

xx

udx

du

du

dy

dx

dy

Page 25: Limits and continuity

25

Higher Derivatives)

If the derivatives of y=f(x) is differentiable function of x, its derivative is called the second derivative of y=f(x) and is denoted by or f ’’(x). That is

Similarly, the third derivative =

the n-th derivative =

2

2

dx

yd

)dx

dy(

dx

d

dx

yd

2

2

)dx

yd(

dx

d

dx

yd)x(f )(

2

2

3

32

)dx

yd(

dx

d

dx

yd)x(f

n

n

n

n)n(

1

1

Page 26: Limits and continuity

26

Example

Find if 3

3

2

2

dx

yd,

dx

yd,

dx

dy 68124 234 xxxxy

8243166

8124 23234

xxxdx

)(d

dx

dx

dx

dx

dx

dx

dx

dx

dx

dy

246488

24316 223

2

2

xxdx

)(d

dx

dx

dx

dx

dx

dx)

dx

dy(

dx

d

dx

yd

69624

6482

2

2

3

3

xdx

)(d

dx

dx

dx

dx)

dx

yd(

dx

d

dx

yd

Page 27: Limits and continuity

27

Differential Calculus

Applications

Page 28: Limits and continuity

28

Slope of a curve

Recall that the derivative of a curve evaluate at a point is the slope of the curve at that point.

f(x0+x)

f(x0)

x0+xx0x

x

y

Y=f(x)

A

B

C

Derivative of f(x) at Xo = slope of f(x) at Xo

Page 29: Limits and continuity

29

Slope of a curve Find the slope of y=2x+3

at x=0 To find the slope of a

curve, we have to compute the derivative of y and then evaluate at a point

The slope of y at x=0 equals 2

(y=mx+c now m=2)

232

dx

)x(d

dx

dy

Page 30: Limits and continuity

30

Slope of a curve Find the slope of at

x=0, 2, -2

The slope of y = 2x The slope of y (at x=0) = 2(0) = 0 The slope of y (at x=-2) =2(-2) = -4 The slope of y (at x=2) =2(2) = 4

xdx

)x(d

dx

dy2

12

12 xy

X=0

X=2X=-2

Page 31: Limits and continuity

31

Local maximum and minimum point

For a continuous function, the point at which is called a stationary point. This gives the point local maximum or local

minimum of the curve

0dx

dy

A

B

C

D

X1

X2

Page 32: Limits and continuity

32

First derivative test (Max)

Given a continuous function y=f(x)

If dy/dx = 0 at x=xo & dy/dx changes from +ve to –ve through x0,

x=x0 is a local maximum point

x=x0

local maximum point

Page 33: Limits and continuity

33

First derivative test (Min)

Given a continuous function y=f(x)

If dy/dx = 0 at x=xo & dy/dx changes from -ve to +ve through x0,

x=x0 is a local minimum point

x=x0 local minimum point

Page 34: Limits and continuity

34

Example Determine the position of any local maximum and

minimum of the function

First, find all stationary point (i.e. find x such that dy/dx = 0)

, so when x=0

By first derivative test x=0 is a local minimum point

12 xy

xdx

)x(d

dx

dy2

12

0dx

dy

020 xdx

dy,xwhen 020 x

dx

dy,xwhen

Page 35: Limits and continuity

35

Example 2

Find the local maximum and minimum of

Find all stationary points first:

296 23 xxxy

296 23 xxxy )x)(x(xxdx

dy3139123 2

310 xandxwhen,dx

dy

x 0 x<1 x=1 1<x<3 x=3 x>3

dy/dx + + 0 - 0 +

Page 36: Limits and continuity

36

Example 2

By first derivative test, x=1 is the local maximum (+ve -> 0 -> -ve) x=3 is the local minimum (-ve -> 0 -> +ve)

Page 37: Limits and continuity

37

Second derivative test

Second derivative test states:

There is a local maximum point in y=f(x) at x=x0, if at x=x0 and < 0 at x=x0.

There is a local minimum point in y=f(x) at x=x0, if at x=x0 and > 0 at x=x0.

If dy/dx = 0 and =0 both at x=x0, the second derivative test fails and we must return to the first derivative test.

0dx

dy2

2

dx

yd

0dx

dy2

2

dx

yd

2

2

dx

yd

Page 38: Limits and continuity

38

Example

Find the local maximum and minimum of

Find all stationary points first:

By second derivative test, x=1 is max and x=3 is min

296 23 xxxy

296 23 xxxy )x)(x(xxdx

dy3139123 2

310 xandxwhen,dx

dyxx

dx

yd126

2

2

0612161

2

2

)(dx

yd

x

0612363

2

2

)(dx

yd

x

Page 39: Limits and continuity

39

x

Page 40: Limits and continuity

40

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41

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42