fundamental engineering maths 2

8/9/2019 Fundamental Engineering Maths 2

1/59

D.T

.PHAM

-VGU

Chapter 2: LIMIT and CONTINUITY

Duong T. PHAM - EEIT2014

Fundamental Engineering Mathematics II for EEIT2014

Vietnamese German UniversityBinh Duong Campus

Duong T. PHAM - EEIT2014 September 30, 2014 1 / 59


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Outline

1 Limit of functions

2 Continuity

3 Derivative and rate of change

4 Derivative as a function



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Limit of a function

Consider f(x) =2x

1 ifx

= 3

6 ifx= 3

x f(x) |5 f(x)|2.99 4.98 0.02

2.999 4.998 0.0022.9999 4.9998 0.0002

2.99999 4.99998 0.00002

2.999999 4.999998 0.000002

2.9999999 4.9999998 0.0000002

x0

y

3

5

5

3

5 +

3+

We say: f converges to5as xgoes to3, and we write

limx3

f(x) = 5



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Limit of a function

Def: Let fbe a function defined on some open intervals that contains a,except possibly at a itself. Then

limxa f(x) =L

if for every >0, there is a number >0 such that

if 0


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Limit of a function

Ex: Prove that limx3

(4x 5) = 7Ans:

Guessing: given an arbitrary >0, we need to find >0 s.t.

if 00, there is =/4 satisfying that if

0


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Limit of a function


(x2 1) = 0Ans:

Guessing: given an arbitrary >0, we need to find >0 s.t.

if 00, there is = min{1/4, /3} satisfying that if0


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Left-hand and right-hand limits

Def: We say thatlim

xaf(x) =L

if for every >0, there is a >0 such that

if a 0 such that

if a


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Ex: Prove that limx0+ xAns:

Guessing: Given >0, we need to find a >0 satisfying

if 00, there is=2 >0such that if 0


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Limit Laws

Limit Laws: Suppose that limxa

f(x) and limxa

g(x) exist and let c

R.

Then

1 limxa

[f(x) +g(x)] = limxa

f(x) + limxa

g(x)

2

limxa[f(x) g(x)] = limxa f(x) limxa g(x)3 lim

xa[cf(x)] =c lim

xaf(x)

4 limxa

[f(x)g(x)] = limxa

f(x) limxa

g(x)

5 limxa

f(x)

g(x)=

limxa

f(x)

limxa

g(x) if lim

xag(x)= 0



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Limit Laws

Proof:



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Some corollaries

In the rest of this course, except when being asked to use the definition of

limit to prove, we can use the six Limit Laws and the following simplelimits without proving:

1 limxa

c=c

2 limxa

x=a

3 limxa

sin x

x = 1

4 limxa

xn =an

5 limxa

n

x= n

a

6 limxa

n

f(x) = n

limxa

f(x) if the second limit exists and ifn is even

we assume further that limxa

f(x)0Proof:



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Limit Laws

Ex: Evaluate limx2

(3x3 2x2 + 10) and limx2

x+ 1

x2

2x+ 3

Ans: limx2

(3x3 2x2 + 10)= limx2

(3x3) limx2

(2x2) + limx2

10

= 3 limx2

x3 2 limx2

x2 + 10

= 3

23

2

22 + 10 =26

and

limx2

x+ 1

x2 2x+ 3 =limx2

(x+ 1)

limx2

(x2 2x+ 3)

=limx2

x+ limx2

1

limx2

x2 2 limx2

x+ limx2

3

= 2 + 1

22 2 2 + 3=1



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Limit Laws

Proposition: If f(x) =g(x)for all x

=a, then lim

xa

f(x) = limxa

g(x)

Ex: Find limx2

f(x) where

f(x) =x2 + 1 if x= 210 if x= 2

Ans: Since f(x) =x2 + 1 for all x= 2, we have

limx2

f(x) = limx2

(x2 + 1)

= 5



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Limit Laws

Ex: Evaluate limx0

(x 2)2 4x

Ans: We have

(x 2)2 4x

=(x 4)xx

=x 4 x= 0.

Hence,

limx0

(x 2)2 4x

= limx0

(x

4) =

4.



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Theorem: limxa

f(x) =L limxaf(x) =Llim

xa+f(x) =L


|x|= 0

Ans: We have|x|=

x ifx0x ifx


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Ex: Prove that limx0 |x

|x does not exist

Ans: We have|x|

x =

x

x = 1 if x>0

xx

=

1 ifx


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Limit of functions

Theorem: Let a(b, c). There holdsf(x)

g(x)

x

(b, c)

\{a

}limxa

f(x) and limxa

g(x) exist

= limxa

f(x) limxa

g(x)

Proof: Denote limxa

f(x) =L and limxa

g(x) =M. Suppose further that M 0.limxa

f(x)=L1 >0 s.t. if 00 s.t. if 0


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Squeeze Theorem

Theorem: Let a

(b, c). There holds

f(x)g(x)h(x) x(b, c)\{a}limxa

f(x) = limxa

g(x) =L

= lim

xag(x) =L

Ex: Evaluate limx0

x2 cos1

x

Ans: We have1cos 1x1 =x2 x2 cos 1

xx2.

Moreover, limx0(x2) = limx0 x2 = 0by Squeeze Theorem, limx0x

2 cos 1x

= 0.


I fi i Li i


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Infinite Limit

Def: Let a (b, c). Then limxa

f(x) = means that for arbitrarypositive number M, there exists >0 satisfying

if 0M

Ex: Prove thatlimx01

|x

|

=.

Ans: Let Mbe an arbitrary positive number. We need to find >0 s.t.

if 0M

But 1|x| >M |x|


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Infinite Limit

Def: Let a (b, c). Then limxa

f(x) = means that for arbitrarynegative number N, there exists >0 satisfying

if 0


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Exercises

2.3:

120; 3438; 46 48;

2.4:

1, 2; 1925; 36, 37, 44.


Continuity


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Continuity

Def: Let a(b, c) and let f be a function defined on (b, c). Functionf is continuous at a if

limxa

f(x) =f(a)

a

f(a)

As xapproaches a

f(x)approachesf(a)


Continuous functions


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Ex: Given f(x) =

1 x ifx1x2 ifx>1.

x

y

1

1

1 2

2

4 f(1) = 1

1 = 0

limx1+

f(x) = limx1+

x2 = 1

limx1

f(x) = limx1+

(1 x) = 0=

limx1

f(x) does not exist

f is NOT continuous at x= 1




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Def: Let f be a function defined on [a, c). Function f is continuousfrom the right at a if

limxa+

f(x) =f(a)

Def: Let f be a function defined on (b, a]. Function f is continuousfrom the left at a if

limxa

f(x) =f(a)




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Ex: Consider again f(x) = 1 x ifx1x2 ifx>1.

x

y

1

1

1 2

2

4

Known: f is NOT continuous atx= 1

limx1+

f(x) = 1; limx1

f(x) = 0;

and f(1) = 0

limx1

f(x) =f(1)

f is continuous from the leftat x= 1

limx1+

f(x)=f(1) f is NOT continuous fromthe right at x= 1


Continuous on intervals


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A function f is said to becontinuous on an open interval(a, b)if itis continuous at any point x(a, b),A function f is said to be continuous on an closed interval [a, b]ifit is continuous at any point x(a, b), and continuous from theright at a and from the left at b.




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Ex: Show that function f(x) = 1 1 x2 is continuous on [1, 1].

Ans:Let1< a< 1. Then

limxa

f(x)= limxa

(1

1 x2) = 1

1 limxa

x2

= 11 a2 =f(a)

= f is continuous at a.Besides,

limx1+

f(x)= limx1+

(11x2) = 11 limx1+ x

2 = 1=f(1)

limx1

f(x)= limx1

(1

1x2) = 1

1 limx1

x2 = 1=f(1)

= f is continuous at1 from the right and at 1 from the left.f is continuous on [1, 1].




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Theorem: If f and g are continuous at x = a. Then the followingfunctions are continuous at x=a:

1 f +g

2 f

g

3 fg

4f

g ifg(a)= 0

5 cf where c is a constant

Proof:




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Theorem:

1 Any polynomial f(x) =anxn +. . .+a1x+a0 is continuous on

(

,

)

2 Any rational function f(x) = g(x)h(x)

, where g and h are polynomials,

is continuous at wherever it is defined.

Proof:




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Ex: Find limx2

x

2

+ 2x 3(x 3)(x+ 3)

Ans:

Function f(x) x2 + 2x 3

(x 3)(x+ 3)is continuous at any point x

=

3 and

x= 3= It is continuous at x=2= lim

x2

x2 + 2x 3(x

3)(x+ 3)

=f(2) =(2)2 + 2 (2) 3

(

2

3)(

2 + 3)

=35




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Theorem: The following types of functions are continuous at every num-ber in their domains:

polynomials

rational functions

root functions

trigonometric functions

inverse trigonometric functions

exponential functions

logarithmic functions


Composition of continuous functions


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p

Theorem: Let f be continuous at a and let g be continuous at f(a).Then, the composition g

f is continuous at a

Proof:


The Intermediate Value Theorem


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Theorem: Let f : [a, b] R be continuous on [a, b]. Let be anynumber between f(a) and f(b), where f(a)

=f(b). Then, there exists a

c(a, b) such that f(c) = .

y=f(x)

a b

f(a)

f(b)

c




34/59

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Ans:

The function f(x) =x3

x+ 2 is continuous on [

2, 0]. Moreover,

f(2) =4 and f(0) = 2.Number 0 satisfies2< 0 < 2. By using Intermediate Value Theorem,there is a c(2, 0)such that f(c) = 0.In other words, the equation: x3 x+ 2 = 0 has a solution c(2, 0).




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y = x3 x+ 2

2

4

2

0c


Exercises


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2.5:

16; 7, 9, 1012, 1520, 3132;

35, 37, 38, 40, 42

4750; 55, 62, 63


Limits at infinity


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x

y

y= 1

x

limx

1

x = ?


Limit at Infinity


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Def: Let fbe a function defined on some interval (a, ). Thenlimx

f(x) =L

if for all >0, there exists a number N such that

if x>N then |f(x) L|<

Ex: Prove that limx

1

x = 0

Ans:

Let >0 be an arbitrary positive number.

We choose N= 1 >0 . Then

ifx>N= 1

x then

1

x

= 1

x0, there exists a number N such that

if xN then f(x)>


Limit at Infinity


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Def: Let fbe a function defined on some interval (a, ). Thenlimx

f(x) =

if for all positive number M, there exists a number N such that

if x>N then f(x)> M

Remark: Note that similar definitions apply when we replaceby.


Horizontal Asymptote


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Def: The line y=L is called a horizontal asymptote of the curve y =

f(x) if eitherlimx

f(x) =L or limx

f(x) =L

Ex: Consider y= x2

1

x2 + 1 .

Since limx

x2 1x2 + 1

=

limx

x2 1

x

2

+ 1

= 1, the

line y = 1 is a horizontalasymptote of the curve.

x

y

1

1


Exercises


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2.6:

1, 3, 4, 11, 13, 14

1530; 3744

53, 55, 57, 61, 68, 69, 71.


Tangent line


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x

y

y = f(x)

Pf(a)

a

t

Qf(x

Q)

xQ

y =f(xQ) f(a)

xQ a(x a) +f(a)

Q

y = limxQa

f(xQ) f(a)

xQ

a

(x a) +f(a)


Tangent line


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Def: The tangent line to the curve y=f(x) at the point P(a, f(a)) isthe line through Pwith slope

m= limxa

f(x) f(a)x a

provided this limit exists.

Ex: Find an equation of the tangent line to the parabola y=x2

at thepoint P(1, 1).

Ans: The slope of the tangent line at the point P(1, 1) is

m= limx1

f(x)

f(1)

x 1 = limx1x2

1

x 1 = limx1(x+ 1) = 2The tangent line is

y 1 = 2(x 1) (y= 2x 1)Duong T. PHAM - EEIT2014 September 30, 2014 44 / 59

Tangent line


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x

y

11 22

1

4

Graph of function f(x) =x2

y= 2x

1


Derivative


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Def: The derivative of a function fat a number x=a, denoted by f(a),is

f(a) = limxa

f(x) f(a)x a

if this limit exists

Remark: The limit in the above definition can be replaced by

f(a) = limh0

f(a+h) f(a)h


Derivative


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Ex: Find the derivative of f(x) =x2 + 2x+ 3at the number x=a

f(a)= limh0

f(a+h) f(a)h

= limh0

(a+h)2 + 2(a+h) + 3

(a2 + 2a+ 3)

h

= limh0

(a+h)2 a2 + 2hh

= limh0

(2a+h)h+ 2h

h

= limh0

(2a+h+ 2)

=2a+ 2


Derivative


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Corollary: The tangent line to to the curvey=f(x)at the point(a, f(a))

is given by y f(a) =f(a)(x a)

Ex: Find an equation of the tangent line to the parabola y=x2 + 2x+ 3at the point (0, 3)

Ans: In the previous example, we have found that

f(a) = 2a+ 2.

Thus, f(0) = 2. Applying the above corollary, the desired tangent line is

y 3 = 2(x 0) or y= 2x+ 3


Rate of change


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Given a function y=f(x), if the variable xchange from x1 to x2,

then the change in x isx= x2 x1

and the correponding change in y is

y= f(x2) f(x1)The difference quotient

y

x

= f(x2) f(x1)

x2

x1

is called the average rate of change ofy with respect to x overthe interval [x1, x2]


Rate of change


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M-VGUThe instantaneous rate of change of y w.r.t. x at x=x1 is

instantaneous rateof change

= limx0

y

x = lim

x2x1

f(x2) f(x1)x2 x1

Note here that

f(x1) = limx0

y

x = lim

x2x1

f(x2) f(x1)x2

x1


Exercises


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2.7:

1, 38, 11, 13, 18, 21, 2530, 3138, 5152


Derivative


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Def: Given f :D R, denote D :={x D such that f(x) exists}.The mapping

f

: D

R

xf(x)is a function ofxand called the derivative of f

Ex: Given f(x) =x3 x. Find f(x).Ans: We have

f(x)= limh0

f(x+h) f(x)h

= limh0

(x+h)3 (x+h) (x3 x)h

= limh0

h[(x+h)2 + (x+h)x+x2] hh

= limh0

(x+h)2 + (x+h)x+x2 1

=3x2 1


Derivative


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x

y

13

13

y = x3 x

x

y

13

13

y = 3x2 1


Differentiable functions


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Def: A function f is differentiable at x=a if f(a)exists.It is differentiable on an interval (a, b)

or (, a) o r (a, ) or

(, ) if it is differentiable at every point in the intervalRemark: The following notations can be used to indicate the thederivative of a function y=f(x)at the number x:

f(x) =y = dy

dx =

df

dx =

d

dxf(x) =Df(x) =Dxf(x)


Differentiable functions


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Ex: Determine when f(x) =|x| is differentiable?

x>0: then f(x) =|x|= xand for sufficiently small|h|, we have x+h>0.Thus

f(x)= limh0

f(x+h)f(x)h

= limh0

|x+h| |x|h

= limh0

x+hxh

= limh0

1= 1

= f is differentiable on(0, )

x


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Ex: Determine when f(x) =|x| is differentiable?

x= 0: then

limh0+

f(0 +h)f(0)h

= limh0+

|0 +h| |0|h

= limh0+

h

h = lim

h0+1= 1

and

limh0f(0 +h)

f(0)

h = limh0 |0 +h

| |0

|h = limh0 h

h = limh0 1=1We note here that

limh0+

f(0 +h)f(0)h

= limh0

f(0 +h)f(0)h

= limh0 f(0+h)f(0)h does not exist f is NOT differentiable at x= 0

Conclusion: f is differentiable in(, 0) (0, ).


Differentiability = continuity?Th If f i diff i bl h f i i


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Theorem: If f is differentiable at a then f is continuous at a

Proof:

f is differentiable at a = limxa

f(x) f(a)x a exists (=L )

Thenlimxa

[f(x) f(a)] = limxa

f(x) f(a)x a (x a)

= lim

xaf(x) f(a)

x a limxa(x a)=L 0 = 0.

Thus, limxa

f(x) = limxa

[f(x) f(a) +f(a)]= lim

xa[f(x) f(a)] + lim

xaf(a)

= 0 +f(a) =f(a).

=

f is continuous at a.


Higher derivatives


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Iffis a differentiable function, thenf is also function. Iff also has a derivative,we then denote f = (f), and f is called thesecond derivative off. We can

write f = ddx

dfdx

= d

2

fdx2

Ex: Given f(x) =x3 x. Find f(x).Ans: On slide 52, we have found that f(x) = 3x2

1. Thus

f(x)= limh0

f(x+h) f(x)h

= limh0

3(x+h)2 1 (3x2 1)h

= limh0

3h(2x+h)

h = lim

h0[3(2x)]

=6x

Def: The third derivativef is defined to be the derivative off, i.e.,f = (f)

and so on ...


Exercises


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2.8:

1, 13, 17, 18, 1924, 31, 4546, 54


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