Title: Functions, Limits and Continuity Prof. Dr. Nasima Akhter And Md. Masum Murshed Lecturer Department of Mathematics, R.U. 29 July, 2011, Friday Time:

Title: Functions, Limits and Continuity

Prof. Dr. Nasima Akhter

And

Md. Masum Murshed

Lecturer

Department of Mathematics, R.U.

29 July, 2011, Friday

Time: 6:00 pm-7:30 pm

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Outline• Functions and its graphs.• One-one, Onto and inverse functions. • Transcendental functions.• Bounded and monotonic functions. • Limits of functions. • Right and left hand limits. • Special limits.• Continuity. • Right and left hand continuity.• Sectional continuity. • Uniform continuity, Lipschitz continuity.

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3

Functions and its Graphs

4

X Y

fx y = f (x)


5

f(x)

x


6

X Yf

x y = f (x)

f : X → Y if for each x ∊ X ∃ a unique y ∊ Y such that y = f(x).


7

f f

f

f is a function

f

f is not a function f is not a function

f is a function f is a function

f


8

X Yf

x y = f (x)


Domain of f

Range of f

Co-domain of f

f[X] = {f(x) : x ∊ X}

= RReal valued function


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X Yf

y = f (x)


Domain of f

Range of f

Co-domain of f

f[X] = {f(x) : x ∊ X}

Set function

Class of sets


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X Yf

y = f (x)

If f : X → Y then ∃ two set functions f : 2X → 2Y and f-1 : 2Y → 2X

Domain of f

Range of f

Co-domain of f

f[A] = {f(x) : x ∊ A}

AA f(A)f(A)


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X Yf

BBf-1(B)f-1(B)f -1(B) = {x ∊ X :

f(x) ∊ B}.

If f : X → Y then ∃ two set functions f : 2X → 2Y and f-1 : 2Y → 2X


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f

f ({1, 3, 4}) = {a, b, d},f ({1, 2}) = {a},

also

f ({2, 3}) = {a, d}

f -1 ({a})= {1, 2}, f -1 ({a, b, d})= {1, 2, 3, 4},f -1 ({b, c})= {4},

f -1 ({c})= Ø,


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Between these two set functions f -1 plays very important role in topology and measure. Since f -1 preserves countable union, countable intersection, difference, monotonicity, complementation etc. i.e.

TheoremIf f : X → Y then for any subset A and B of Y,

(i) f -1(A ∪ B) = f -1(A) ∪ f -1(B).(ii) f -1(A ∩ B) = f -1(A) ∩ f -1(B).(iii) f -1(A ∖ B) = f -1(A) ∖ f -1(B). (iv) If A ⊆ B then f -1(A) ⊆ f -1(B).(v) f -1(Ac) = (f -1(A))c.And, more generally, for any indexed {Ai} of subsets of Y,(vi) f -1( ∪i Ai) = ∪i f -1(Ai).(vii) f -1(∩i Ai) = ∩i f -1(Ai).


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xx yy

X Y

X ⨉ Y = {(x, y) : x ∊ X , y ∊ Y}

(x, y)

A relation from X to Y is a subset of X ⨉ Y.


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A function can also be described as a set of ordered pairs (x, y) such that for any x-value in the set, there is only one y-value. This means that there cannot be any repeated x-values with different y-values.

A relation is called a function if for any x-value in the set, there is only one y-value. This means that there cannot be any repeated x-values with different y-values.

f

f is not a function

g

g is a function

f = {(1, a), (1, b), (2, c), (3, c), (4, d)} is not a function.

g = {(1, a), (2, a), (3, d), (4, b)} is a function.


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y = f(x)y = f(x)

X, A subset of R

We plot the domain X on the x-axis, and the co-domain Y on the y-axis. Then for each point x in X we plot the point (x, y), where y = f(x). The totality of such points (x, y) is called the graph of the function.

(x, y)

xxf

Y, A subset of R


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This is the graph of a function. All possible vertical lines will cut this graph only once.

This is not the graph of a function. The vertical line we have drawn cuts the graph twice.

The Vertical Line Test

Functions and its GraphsNow we consider some examples of real functions. Example 1. The identity function. Let the function f : R → R be defined by f(x) = x for all real x. This function is often called

the identity function on R and it is denoted by 1R. Its domain is the real line, that is, the set of all real numbers. Here x = y for each point (x, y) on the graph of f. The graph is a straight line making equal angles with the coordinates axes (see Figure-1 ). The range of f is the set of all real numbers.

Figure-1 Graph of the identity function f(x) = x. 18

Functions and its GraphsExample 2.

The absolute-value function.

Consider the function which assigns to each real number x the nonnegative number |x|. We define the function y = |x| as

From this definition we can graph the function by taking each part separately. The graph of y = |x| is given below.

Figure- 2 Graph of the absolute-value function y = f(x) = |x|. 19


Constant functions.

A function whose range consists of a single number is called a constant function. An example is shown in Figure-3, where f(x) = 3 for every real x. The graph is a horizontal line cutting the Y-axis at the point (0, 3).

Figure- 3 Graph of the constant function f(x) = 3.

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Linear functions and affine linear function.

Let the function g be defined for all real x by a formula of the form g(x) = ax + b, where a and b are real numbers, then g is called a linear function if b = 0. Otherwise, g is called a affine linear function. The example, f(x) = x, shown in Figure-1 is a linear function. And, g(x) = 2x - 1, shown in Figure-4 is a affine linear function.

Figure- 4 Graph of the affine linear function g(x) = 2x - 1. 21


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Example 5.The greatest integer function The greatest integer function is defined by f(x) = [x] = The greatest integer less than or equal to x. Figure- 5 shows the graph of f(x) = [x].

Figure- 5 Graph of the greatest integer function is defined by f(x) = [x].


Polynomial functions.

A polynomial function P is one defined for all real a by an equation of the form

P(x) = c0 + c1x + c2x2 + c3x3 + ……………+ cnxn

The numbers c0 , c1, c2, c3,……………,cn are called the coefficients of the polynomial,

and the nonnegative integer n is called its degree (if cn ≠ 0). They include the constant functions and the power functions as special cases. Polynomials of degree 2, 3, and 4 are called quadratic, cubic, and quartic polynomials, respectively. Figure-6 shows a portion of the graph of a quartic polynomial P given by P(x) = ½ x4 – 2x2.

Figure- 6 Graph of a quartic polynomial function p(x) = ½ x4 – 2x2.

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24

f g

are functions

h1 = f +g

h2 = f .g

h3 = f /g

h7 = g∘ f h6 = f ∘g

k = constant

h4 = k + f

h5 = k . f


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f and g are real valued and defined on the same domain X and k is a real number.

if

h1 = f +g, h2 = f .g , h3 = f /g,

h7 = g∘ f h6 = f ∘g and

We can define

h1(x)= (f +g)(x) = f(x) + g(x),

by

h2(x)= (f .g)(x) = f(x) . g(x),

h3(x)= (f /g)(x) = f(x) / g(x),

Also we can define

byh6(x)= (f ∘g)(x) = f(g(x)) if co-domain of g = domain of f

and h7(x)= (g ∘f)(x) = g (f(x)) if co-domain of f = domain of g

Generally, f ∘g ≠ g∘ f

h4 = k + f and h5 = k . f

h4(x)= (k + f)(x) = k + f(x) , h5(x)= (k . f)(x) = k . f(x) ,

Composition Function

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If f : X → Y and g : Y → Z then we define a function (g∘ f) : X → Z by (g∘ f)(x) ≡ g(f(x)).

Composition FunctionExample

Let f : X → Y and g : Y → Z be define by the following diagrams

Then we can compute (g∘ f) : X → Z by its definition:

(g∘ f)(a) ≡ g(f(a)) = g(y) = t

(g∘ f)(b) ≡ g(f(b)) = g(z) = r

(g∘ f)(c) ≡ g(f(c)) = g(y) = t

Remark:

Let f : X → Y. Then 1Y∘ f = f and f∘ 1X = f where 1X and 1Y are identity functions on X and Y respectively. That is the product of any function and the identity function is the function itself.

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Theorem Two functions f and g are equal if and only if (a) f and g have the same domain, and (b) f(x) = g(x) for every x in the domain of f.

Theorem: Let f : X → Y, g : Y → Z and h : Z → W. Then (h∘ g)∘ f = h∘ (g∘ f).

Outline• Functions and its graphs.• One-one, Onto and inverse functions. • Transcendental functions. • Bounded and monotonic functions. • Limits of functions. • Right and left hand limits. • Special limits.• Continuity. • Right and left hand continuity.• Sectional continuity. • Uniform continuity, Semi-continuity, Lipschitz

continuity. 29

One-one Function

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Different elements

of X

Different elements

of X

Different elements of Y

Different elements of Y

X Y

f is one-one

f : X → Y is said to be one-one if f(x) = f(y) implies x = y or, equivalently, x ≠ y implies f(x) ≠ f(y).

One-one FunctionExamples

(vi) Let X = {1, 2, 3, 4} and Y = {a, b, c, d, e} and f and g be two functions from X into Y given by the following diagrams

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Here f is not a one-one function since a is the image of two different elements 1 and 2 of X.

Here g is a one-one function since different elements of X have different images.


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This is the graph of a one- one function. All possible horizontal lines will cut this graph only once.

This is not the graph of a one- one function. The horizontal line we have drawn cuts the graph twice.

The Horizontal Line Test

One-one Function

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Examples

The function f : R → R defined by f(x) = x2 is not a one-one function since f(2) = f(-2) = 4.

Figure- 7 Graph of the function f(x) = x2.

One-one Function

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Examples

The function f : R → R defined by f(x) = ex is a one-one function.

Figure- 8 Graph of the function f(x) = ex.

Proof:Let f(x) = f(y) then ex = ey i.e. x = yHence by definition f is a one-one function.

One-one Function

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Examples

The absolute-value function f : R → R defined by f(x) =|x |is not a one-one function since f(2) = f(-2) = 2.

Figure- 9 Graph of the absolute-value function y = f(x) = |x|.

One-one Function

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Examples

The identity function f : R → R defined by f(x) = x is a one-one function.

Figure- 10 Graph of the identity function f(x) = x.

Proof:Let f(x) = f(y) then x = yHence by definition f is a one-one function.

Onto Function

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Every element of Y is the

image of some element of X

Every element of Y is the

image of some element of X

X Y

f is onto

f : X → Y is said to be onto if for every y ϵ Y, ∃ an element x ϵ X such that y = f(x), i.e. f(X) = Y.

Onto FunctionExamples

(vi) Let X = A = {1, 2, 3, 4}, Y = {a, b, c} and B= {a, b, c, d, e} and f and g be two functions from X into Y and A into B respectively given by the following diagrams

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Here f is a onto function since every element of Y appears in the range of f

Here g is not a onto function since e ∊ B is not an image of any element of A.

Onto Function

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This is the graph of a onto function. All possible horizontal lines will cut this graph at least once.

This is not the graph of a onto function. the horizontal line drawn above does not cut the graph.

The Horizontal Line Test

Onto Function

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Examples

The function f : R → R defined by f(x) = ex is a not an onto function.

Figure- 11 Graph of the function f(x) = ex.

Proof:Since -2 is an element of the co-domain R then there does not exist any element x in the domain R such that -2 = ex = f(x). Hence by definition f is not a onto function.

Onto Function

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Examples

The identity function f : R → R defined by f(x) = x is a onto function.

Figure- 12 Graph of the identity function f(x) = x.

Proof:Since for every y in the co-domain R, ∃ an element x in the domain R such that y = f(x). Hence by definition f is a onto function.

Onto Function

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Examples

The function f : R → R defined by f(x) = x2 is not an onto function.

Figure- 13 Graph of the function f(x) = x2.

Proof:Since -2 is an element of the co-domain R then there does not exist any element x in the domain R such that -2 = x2 = f(x). Hence by definition f is not a onto function.

Inverse Function

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X Yf is one-one and onto

x y = f(x)

f

f -1

Inverse Function

44

YX

Y X

Example Let f : X → Y be define by the following diagram

Here f is one-one and onto. Therefore f -1, the inverse function, exists.We describe f -1 : Y → X by the diagram

Inverse FunctionTheorem on the inverse function.

Theorem 1

Let the function f : X → Y be one-one and onto; i.e. the inverse function f -1 : Y → X exists. Then the product function

(f -1 ∘ f ) : X → X

is the identity function on X, and the product function

(f ∘ f -1 ) : Y → Y

is the identity function on Y, i.e. (f -1 ∘ f ) = 1X and (f ∘ f -1 ) = 1Y.

Theorem 2

Let f : X → Y and g : Y → X. Then g is the inverse of f, i.e. g = f -1 , if the product function

(g ∘ f ) : X → X

is the identity function on X, and the product function

(f ∘ g) : Y → Y

is the identity function on Y, i.e. (g ∘ f ) = 1X and (f ∘ g ) = 1Y.

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Algebraic and Transcendental Functions

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Algebraic functions Algebraic functions are functions y = f (x) satisfying an equation of the formp0(x)yn + p1(x)yn-1 + . . . + pn-1(x)y + pn(x) = 0 …………………………… (1) where p0(x) , . . . , pn(x) are polynomials in x.

If the function can be expressed as the quotient of two polynomials, i.e., P(x)/Q(x) where P(x) and Q(x) are polynomials, it is called a rational algebraic function; otherwise, it is an irrational algebraic function.

Transcendental functions Transcendental functions are functions which are not algebraic; i.e., they do not satisfy equations of the form of Equation (1).

Example is an algebraic function since it satisfies the equation(x2 – 2x + 1)y2 + (2x2 – 2x)y + (x2 – x) = 0.

1

x

xxy

Algebraic and Transcendental Functions

48

The following are sometimes called elementary transcendental functions.

1. Exponential function: f (x) = ax, a ≠ 0, 1.

2. Logarithmic function: f (x) = logax, a ≠ 0, 1. This and the exponential function are inverse functions.If a = e = 2.71828 . . . , called the natural base of logarithms,

we write f (x) = logex = In x, called the natural logarithm of x.3. Trigonometric functions:

The variable x is generally expressed in radians (π radians = 180∘). For real values of x, sin x and cos x lie between –1 and 1 inclusive.4. Inverse trigonometric functions. The following is a list of the inverse trigonometric functions and their principal values:

Outline• Functions and its graphs.• One-one, Onto and inverse functions. • Transcendental functions. • Bounded and monotonic functions. • Limits of functions. • Right and left hand limits. • Special limits.• Continuity. • Right and left hand continuity.• Sectional continuity. • Uniform continuity, Lipschitz continuity.

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Bounded FunctionBounded functionA function f defined on some set X with real values is called bounded, if the set of its

values is bounded. In other words, there exists a real number M < ∞ such that |f(x)| ≤ M or –M ≤ f(x) ≤ M for all x in X.

Geometrically, the graph of such a function lies between the graphs of two constant step functions s and t having the values — M and +M, respectively.

Figure- 14 Graph of a bounded function.

50Intuitively, the graph of a bounded function stays within a horizontal band.

Bounded Function

Figure- 16 Graph of a bounded below function.

51

Bounded above function and Bounded below functionSometimes, if f(x) ≤ A for all x in X, then the function is said to be bounded above by A. On the other hand, if f(x) ≥ B for all x in X, then the function is said to be bounded below by B.

Figure- 15 Graph of a bounded above function.

Bounded Function

52

Unbounded functionThe function which is not bounded is called unbounded function.

Figure- 17 Graphs of un-bounded functions.

Monotonic FunctionMonotonic functions

A function f is said to be increasing on a set S if f(x) ≤ f(y) for every pair of points x

and y in S with x ≤ y. If the strict inequality f(x) < f(y) holds for all x < y in S, the

function is said to be strictly increasing on S.

Similarly, f is called decreasing on S if f(x) ≥ f(y) for all x ≤ y in S. If f(x) > f(y) for all x < y in S, then f is called strictly decreasing on S.

A function is called monotonic on S if it is increasing on S or if it is decreasing on S.

The term strictly monotonic means that is strictly increasing on S or strictly decreasing on S. Ordinarily, the set S under consideration is either an open interval or a closed interval.

Figure- 18 Monotonic functions. 53

Monotonic FunctionPiecewise Monotonic functions

A function f is said to be piecewise monotonic on an interval if its graph consists of a finite number of monotonic pieces.

That is to say, f is piecewise monotonic on [a, b] if there is a partition P of [a, b] such that f is monotonic on each of the open subintervals of P.

In particular, step functions are piecewise monotonic, as are all the examples shown in Figures- 23 and 24.

Figure- 19 Piecewise Monotonic function. 54

Monotonic FunctionExample 1.

The power functions.

If p is a positive integer, we have the inequality xp < yp if 0 ≤ x < y, which is easily proved by mathematical induction. This shows that the power function f, defined for all real x by the equation f(x) = xp, is strictly increasing on the nonnegative real axis. It is also strictly monotonic on the negative real axis (it is decreasing if p is even and increasing if p is odd). Therefore, f is piecewise monotonic on every finite interval.

Example 2.

The square-root function.

Let f(x) = √x for x > 0. This function is strictly increasing on the nonnegative real axis. In fact, if 0 ≤ x < y, we have

Example 3.

The graph of the function g defined by the equation g(x) = √(r2-x2) if — r ≤ x ≤ r

is a semicircle of radius r. This function is strictly increasing on the interval — r ≤ x ≤ 0

and strictly decreasing on the interval 0 ≤ x ≤ r. Hence, g is piecewise monotonic on

[-r, r].

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Outline• Functions and its graphs.• One-one, Onto and inverse functions. • Principal values.• Transcendental functions. • Bounded and monotonic functions. • Limits of functions. • Right and left hand limits. • Special limits.• Continuity. • Right and left hand continuity.• Sectional continuity. • Uniform continuity, Lipschitz continuity. 56

Limit of a Function

57

Example

Limit of a Function

58

Example

Limit of a Function

59

Figure- 20

Limit of a Function

60

Example

Figure- 21

Limit of a Function

61

Example

Figure- 22

Limit of a Function

62

Example

Figure- 23

Limit of a Function

63

Neighborhood of a point Any open interval containing a point p as its midpoint is called a neighborhood of p. The neighborhood of p with radius r is denoted by Nr(p) and defined by Nr(p) = {x ϵ R : |x - p| < r} = (p – r, p + r).

Deleted neighborhood of a point Nr*(p) = Nr(p) \ {p} is called the deleted neighborhood of p, i.e. Nr*(p) = {x ϵ R : x ≠ p and |x - p| < r}.

Limit of a Function

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Or equivalently

if for every neighborhood Nε(L) there is some deleted neighborhood Nδ(a) such that if x ϵ Nδ(a) then f(x) ϵ Nε(L), i.e. for every neighborhood Nε(L) there is some neighborhood Nδ(a) such that if x ϵ Nδ(a) \ {a} then f(x) ϵ Nε(L).

Or equivalently

if for every neighborhood Nε(L) there is some deleted neighborhood Nδ(a) such that if x ϵ Nδ(a) then f(x) ϵ Nε(L), i.e. for every neighborhood Nε(L) there is some neighborhood Nδ(a) such that if x ϵ Nδ(a) \ {a} then f(x) ϵ Nε(L).

Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a . Then we say that,

if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever 0 < |x-a| < δ


if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever 0 < |x-a| < δ

Limit of a Function

65

Figure- 24

f(x)

L

L - ε

L + ε

a a - δ a + δ

Limit of a FunctionFind the value of delta () and Epsilon ()

66Figure- 25

Limit of a Function

67

Procedure for finding limitProcedure for finding limit

Step-1 Guess LGuess L

Step-2 Choose an arbitrary (small) number ε > 0.Choose an arbitrary (small) number ε > 0.

Step-3 Find the value of δ > 0 which satisfies the following conditionFind the value of δ > 0 which satisfies the following condition

|f(x) - L|< ε whenever 0 < |x-a| < δ |f(x) - L|< ε whenever 0 < |x-a| < δ

Limit of a FunctionExample Use the definition of the limit to prove the following limit.

Solution:Let ε > 0 be any arbitrary (small) number then we need to find a number δ > 0 so that

the following will be true. |(5x - 4) - 6|< ε whenever 0 < |x-2| < δSimplifying the left inequality in an attempt to get a guess for δ, we get,|(5x - 4) - 6| = |5x - 10| = 5|x - 2| < ε ⇒ |x - 2| < ε/5. So, it looks like if we choose δ = ε/5 we should get what we want.Let’s now verify this guess. So, again let ε > 0 be any number and then choose δ = ε/5.Now, assume 0 < |x-2| < δ = ε/5 then we get,|(5x - 4) - 6| = |5x - 10| = 5|x - 2| < 5(ε/5) = ε.So, we’ve shown that |(5x - 4) - 6|< ε whenever 0 < |x-2| < δ = ε/5 Hence by definition, 68

Limit of a Function

69

Limit of a Function

70

ExampleUse the definition of the limit to prove the following limit.

Solution:Let ε > 0 be any arbitrary (small) number then we need to find a number δ > 0 so that the following will be true.

| - 4| < ε whenever 0 < |x-2| < δ

Since x ≠ 2 then for all , f(x) =

Assume δ ≤ 1, then |x-2| < δ implies 1 < x < 3. Thus

So, we can choose δ = ε.

Then we have, |f(x) - 4|< ε whenever 0 < |x-2| < δ

Hence by definition,

Limit of a Function

71

Figure- 26

4 - ε

4 + ε

2 - δ 2 + δ

Limit of a Function

72

Solution:Let ε > 0 be any arbitrary (small) number then we need to find a number δ > 0 so that the following will be true.

|f(x) - 0|< ε whenever 0 < |x-0| < δ

Simplifying the left inequality in an attempt to get a guess for δ, we get,

= since 1

So, we can choose δ = ε.

Then we have, |f(x) - 0|< ε whenever 0 < |x-0| < δ

Hence by definition,

Limit of a Function

73

Theorem

Proof


Right hand limit

75

Or equivalently

if for every neighborhood Nε(L) there is some number δ > 0 such that if x ϵ (a, a + δ) then f(x) ϵ Nε(L).

Or equivalently

if for every neighborhood Nε(L) there is some number δ > 0 such that if x ϵ (a, a + δ) then f(x) ϵ Nε(L).

Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a . Then for right hand limit we say,

if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever 0 < x-a < δ (Or a < x < a + δ)

Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a . Then for right hand limit we say,

if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever 0 < x-a < δ (Or a < x < a + δ)

Right hand limit

76

Figure- 27

f(x)

L

L - ε

L + ε

a a + δ

Right hand limit Example Use the definition of the limit to prove the following limit.

SolutionLet ε > 0 be any arbitrary (small) number then we need to find a number δ > 0 so that

the following will be true. |√x - 0|< ε whenever 0 < x-0 < δi.e. we need to show that|√x|< ε whenever 0 < x < δSimplifying the left inequality in an attempt to get a guess for δ, we get,|√x|< ε ⇒ x < ε2. So, it looks like we can choose δ = ε2.Let’s now verify this guess. So, again let ε > 0 be any number and then choose δ = ε2.Now, assume 0 < x- 0 < δ = ε2 then we get,|√x - 0| = |√x| < | √ ε2 | = ε.So, we’ve shown that |√x - 0|< ε whenever 0 < x-0 < δ = ε2

Hence by definition, 77

Left hand limit

78

Or equivalently

if for every neighborhood Nε(L) there is some number δ > 0 such that if x ϵ (a - δ, a ) then f(x) ϵ Nε(L).

Or equivalently

if for every neighborhood Nε(L) there is some number δ > 0 such that if x ϵ (a - δ, a ) then f(x) ϵ Nε(L).

Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a . Then for left hand limit we say,

if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever – δ < x-a < 0 (Or a - δ < x < a)

Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a . Then for left hand limit we say,

if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever – δ < x-a < 0 (Or a - δ < x < a)

Left hand limit

79

Figure- 28

f(x)

L

L - ε

L + ε

aa - δ

Right and left hand limits

80

Figure- 29

4 - ε

4 + ε

2 - δ 2 + δ

=


81

Example

Figure- 30

x

f(x)

1

-1

Right and left hand limits.

82

That is the definition of left hand limit is satisfied for δ = ε.


83Figure- 31 Figure- 32


84

L

+

-

Figure- 33


85

f(x)

Figure- 34


. toequal and

exist both )(lim and )(lim

ifonly and if exists )(lim

-

L

xfxf

Lxf

axax

ax

86

Limit Properties

87

Limit Properties

88

Using the limit properties We can easily find the following limits

Limit Properties

89

We can take this fact one step farther to get the following theorem.

Limit Properties

90

Figure- 35

The following figure illustrates what is happening in Squeeze Theorem.

The Squeeze theorem is also known as the Sandwich Theorem and the Pinching Theorem.


Special Limits

92


if for every number M > 0 there is some number δ > 0 (usually depending on M and a) such that f(x) > M whenever 0 < |x-a| < δ


if for every number M > 0 there is some number δ > 0 (usually depending on M and a) such that f(x) > M whenever 0 < |x-a| < δ

Figure- 36

Special Limits

93

Example Use the definition of the limit to prove the following limit.

SolutionLet M > 0 be any number and we’ll need to choose a δ > 0 so that,

i.e.

Special Limits

94

Thus we have

Special Limits

95Figure- 37

aa - δ a + δ

N

f(x)


if for every number N < 0 there is some number δ > 0 (usually depending on N and a) such that f(x) < N whenever 0 < |x-a| < δ


if for every number N < 0 there is some number δ > 0 (usually depending on N and a) such that f(x) < N whenever 0 < |x-a| < δ

Special Limits

96Figure- 38


if for every number ε > 0 (however small) there is some number M > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever x > M.


if for every number ε > 0 (however small) there is some number M > 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever x > M.

Special Limits

,0lime.g.

x

xe

97

Figure- 39

Special Limits

98Figure- 40


if for every number ε > 0 (however small) there is some number N < 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever x < N.


if for every number ε > 0 (however small) there is some number N < 0 (usually depending on ε and a) such that | f(x) - L| < ε whenever x < N.

f(x)

N x

y

L + ε

LL - ε

Special Limits

99


if for every number N > 0 there is some number M > 0 (usually depending on N ) such that f(x) > N whenever x > M.


if for every number N > 0 there is some number M > 0 (usually depending on N ) such that f(x) > N whenever x > M.

N

M

y

x

f(x)

Figure- 41

Special Limits

100

If

wE say that limit of f(x) does not exist as x tends to a. These limits exist in the extended real line.

If

wE say that limit of f(x) does not exist as x tends to a. These limits exist in the extended real line.

Special LimitsProperties of limit of functions :

.)(

1limthen ,0)(and0)(lim If )1(

00

xf

xfxfxxxx

.0)(

1limthen ),or()(lim If )2(

00

xf

xfxxxx

)..resp()]()([lim

then ), resp.()(lim)(lim If )3(

0

00

xgxf

xgxf

xx

xxxx

101

Special LimitsProperties of limit of functions :

.constant realany for

])([limthen ,)(lim If )4(00

k

kxfxfxxxx

.constant zero-nonany for

)]([limthen ,)(lim If )5(00

k

xkfxfxxxx

.)]()([lim

then ,)(lim)(lim If )6(

0

00

xgxf

xgxf

xx

xxxx

102

Special Limits

103

Special Limits

104Figure- 42

Present a geometric proof of the following limit.

ProofConstruct a circle with center at O and radius OA = OD = 1, as in Figure 33. Choose point B on OA extended and point C on OD so that lines BD and AC ar perpendicular to OD. It is geometrically evident that Area of triangle OAC < Area of sector OAD < Area of triangle OBD that is,


Continuous Function

106

A function f(x) is said to be continuous x = a, if for every neighborhood Nε(f(a)) there is some neighborhood Nδ(a) such that if x ϵ Nδ(a) then f(x) ϵ Nε(f(a)).

A function f(x) is said to be continuous x = a, if for every neighborhood Nε(f(a)) there is some neighborhood Nδ(a) such that if x ϵ Nδ(a) then f(x) ϵ Nε(f(a)).

A function f(x) is said to be continuous x = a, if

(i) f(x) is well defined at x = a, and

(ii)

A function f(x) is said to be continuous x = a, if

(i) f(x) is well defined at x = a, and

(ii)

A function f(x) is said to be continuous x = a, if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) – f(a)| < ε whenever |x-a| < δ

A function f(x) is said to be continuous x = a, if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) – f(a)| < ε whenever |x-a| < δ

Continuous Function

107

Figure- 43

f(x)

f(a)

f(a) - ε

f(a) + ε

a a - δ a + δ

Continuous Function

).,( allfor ),()(lim i.e.

),,(

in point evey at continuous is)(

ifonly and if ),( intervalopen

in the continuous be tosaid is )(

baccfxf

ba

xf

ba

xf

cx

108

Continuous Function

.)()(lim and ),()(lim

and ),,(in continuous is )(

ifonly and if ],[ interval closed

on the continuous be tosaid is )(

bfxfafxf

baxf

ba

xf

bxax

109

Continuous Function

110

Figure- 44 Figure- 45


Right hand continuity

112

Or equivalently if for every neighborhood Nε(f(a)) there is some number δ > 0 such that if x ϵ [a, a + δ) then f(x) ϵ Nε(f(a)).

Or equivalently if for every neighborhood Nε(f(a)) there is some number δ > 0 such that if x ϵ [a, a + δ) then f(x) ϵ Nε(f(a)).

Let f(x) be a function defined on an interval that contains x = a. Then f(x) is said to be right hand continuous or right continuous at x = a, if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - f(a) | < ε whenever 0 ≤ x-a < δ (Or a ≤ x < a + δ)

Let f(x) be a function defined on an interval that contains x = a. Then f(x) is said to be right hand continuous or right continuous at x = a, if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - f(a) | < ε whenever 0 ≤ x-a < δ (Or a ≤ x < a + δ)

. at continuousright

be tosaid is )(),()(limIf

ax

xfafxfax


113

Figure- 46

f(x)

f(a)

f(a) - ε

f(a) + ε

a a + δ


114

Example

Figure- 47

H(t) is right continuous at t = 0.

Left hand continuity

115

Or equivalently if for every neighborhood Nε(f(a)) there is some number δ > 0 such that if x ϵ (a - δ, a ] then f(x) ϵ Nε(f(a)).

Or equivalently if for every neighborhood Nε(f(a)) there is some number δ > 0 such that if x ϵ (a - δ, a ] then f(x) ϵ Nε(f(a)).

Let f(x) be a function defined on an interval that contains x = a. Then f(x) is said to be left hand continuous or left continuous at x = a, if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - f(a) | < ε whenever – δ < x-a ≤ 0 (Or a - δ < x ≤ a)

Let f(x) be a function defined on an interval that contains x = a. Then f(x) is said to be left hand continuous or left continuous at x = a, if for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) - f(a) | < ε whenever – δ < x-a ≤ 0 (Or a - δ < x ≤ a)

. at continuousleft

be tosaid is )(),()(limIf

ax

xfafxfax


116

Figure- 48

f(x)

f(a)

f(a) - ε

f(a) + ε

aa - δ


117

Figure- 49

f(a) f(x) is left continuous at x = a.

Continuous Functions

118

(1) f(x) is defined at all points inside a neighborhood of the point a.

(2) Left hand limit and right hand limit of f(x) exist at x = a and both have equal values, i.e. f(a+) = f(a-) = L, Where f(a+) and f(a-) denote the left hand limit and the right hand limit of f(x) at x = a respectively.

(3) The value of f(x) at x = a is equal to L, i.e. f(a) = L.

To show continuity of f (x) at x = a, the following conditions must be verified:

If any of the above conditions is violated, then f(x) is said to be discontinuous at x = a.


119

There are two kinds of discontinuity.

(1) f(x) has discontinuity of the first kind at x = a if f(a+) and f(a-) exist, but at least one of them is different from f(a).

(2) f(x) has discontinuity of the second kind at x = a if at least one of f(a+) and f(a-) does not exist.

There are two types of the Discontinuity of the first kind.

(i) Solvable discontinuity: if f(a+) = f(a-).

(ii) Jump discontinuity: if f(a+) ≠ f(a-).


120Figure- 50

f(2) = 4

Solvable discontinuity


121

Example

Figure- 51

Jump discontinuity


122

Example

Figure- 52

Jump discontinuity


123

Continuous FunctionsExampleDirichlet Function

The Dirichlet Function d : R → R is defined as follows:

d(x) = 1 if x is rational 0 if x is irrational

0

Y

X

1

Figure-53 Graph of the function d(x) = 1 if x is rational 0 if x is irrational 124

Discontinuity of the second kind

Continuous FunctionsExampleModification-1 of Dirichlet Function

The Modification-1 of Dirichlet Function f : R → R is defined as follows:

f(x) = x. d(x) x if x is rational 0 if x is irrational

Figure-54 Graph of the function f(x) = x if x is rational 0 if x is irrational125

This function is continuous at only x = 0.

Continuous FunctionsExampleModification-2 of Dirichlet Function


f(x) = x2. d(x) x2 if x is rational 0 if x is irrational

Figure-55 Graph of the function

f(x) = x2 if x is rational 0 if x is irrational 126

This function is continuous at only x = 0.

Continuous FunctionsExampleModification-3 of Dirichlet Function. It is also known as ruler function.


Figure-56 Graph of the ruler function. 127

This function is continuous at every irrational numbers.


128


129


Sectional continuity or piecewise continuity

131

Figure- 57


Uniform Continuity

133

A function f(x) is said to be uniformly continuous in an interval if for every number ε > 0 (however small) there is some number δ > 0 (depending on ε ) such that | f(x1) – f(x2)| < ε whenever |x1-x2| < δ, where x1and x2 are any two points in the interval.

A function f(x) is said to be uniformly continuous in an interval if for every number ε > 0 (however small) there is some number δ > 0 (depending on ε ) such that | f(x1) – f(x2)| < ε whenever |x1-x2| < δ, where x1and x2 are any two points in the interval.

Let f(x) be continuous in an interval. Then, by definition, at each point a of the interval and for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) – f(a)| < ε whenever |x-a| < δ. If we can find δ for each ε which holds for all points of the interval (i.e. δ depends only one ε and not on a) we say that is uniformly continuous in the interval.

Let f(x) be continuous in an interval. Then, by definition, at each point a of the interval and for every number ε > 0 (however small) there is some number δ > 0 (usually depending on ε and a) such that | f(x) – f(a)| < ε whenever |x-a| < δ. If we can find δ for each ε which holds for all points of the interval (i.e. δ depends only one ε and not on a) we say that is uniformly continuous in the interval.

Uniform Continuity

134

Figure- 58

f(x)

f(x1)ε

δ

x2x1

f(x2)

a b

Uniform Continuity

135

ProblemProve that f (x) = x2 is uniformly continuous in 0 < x < 1.ProblemProve that f (x) = x2 is uniformly continuous in 0 < x < 1.

Theorem If f is continuous in a closed interval, it is uniformly continuous in the interval.

Theorem If f is continuous in a closed interval, it is uniformly continuous in the interval.

SolutionWe must show that, given any ε > 0, we can find δ > 0 such that | x1

2– x22| < ε whenever |x1-x2| < δ, where x1, x2 ϵ (0, 1).Now

| x12– x2

2)| = |x1+ x2| |x1-x2| ≤ (|x1| + |x2|) < (1 + 1) |x1-x2| =2 |x1-x2|

Therefore given ε > 0 let δ = ε /2. Then if |x1-x2| < δ, then | x12– x2

2| < ε .

SolutionWe must show that, given any ε > 0, we can find δ > 0 such that | x1

2– x22| < ε whenever |x1-x2| < δ, where x1, x2 ϵ (0, 1).Now

| x12– x2

2)| = |x1+ x2| |x1-x2| ≤ (|x1| + |x2|) < (1 + 1) |x1-x2| =2 |x1-x2|

Therefore given ε > 0 let δ = ε /2. Then if |x1-x2| < δ, then | x12– x2

2| < ε .

Uniform Continuity

136

ProblemProve that f (x) = x2 is not uniformly continuous in R = (-∞, ∞).ProblemProve that f (x) = x2 is not uniformly continuous in R = (-∞, ∞).

SolutionSuppose that f (x) = x2 is uniformly continuous in R, then for all ε > 0, there would exist a δ > 0 such that | x1

2– x22| < ε whenever |x1-x2| < δ

where x1, x2 ϵ (0, 1). Take x1 > 0 and let x2 = x1 + δ/2. Write ε ≥ | x1

2– x22)| = |x1+ x2| |x1-x2| = (2x1 + δ/2)δ/2 > x1δ

Therefore x1 ≤ ε/δ for all x1 > 0, which is a contradiction. It follows that f (x) = x2 cannot be uniformly continuous in R = (-∞, ∞).

SolutionSuppose that f (x) = x2 is uniformly continuous in R, then for all ε > 0, there would exist a δ > 0 such that | x1

2– x22| < ε whenever |x1-x2| < δ

where x1, x2 ϵ (0, 1). Take x1 > 0 and let x2 = x1 + δ/2. Write ε ≥ | x1

2– x22)| = |x1+ x2| |x1-x2| = (2x1 + δ/2)δ/2 > x1δ

Therefore x1 ≤ ε/δ for all x1 > 0, which is a contradiction. It follows that f (x) = x2 cannot be uniformly continuous in R = (-∞, ∞).

Uniform Continuity

137

ProblemProve that f (x) = 1/x is not uniformly continuous in 0 < x < 1.ProblemProve that f (x) = 1/x is not uniformly continuous in 0 < x < 1.

SolutionSuppose that f (x) = 1/x is uniformly continuous in R, then for all ε > 0, there would exist a δ > 0 such that | 1/x1

– 1/x2| < ε whenever |x1-x2| < δ where x1, x2 ϵ (0, 1). ε > | 1/x1

– 1/x2| = (|x2-x1| )/ (x1x2)Or |x1-x2| < x1x2 εTherefore, to satisfy the definition of uniform continuity we would have to have δ ≤ x1x2 ε for all x1, x2 ϵ (0, 1), but that would mean that δ ≤ 0. Therefore there is no single δ > 0. Therefore f (x) = 1/x is not uniformly continuous in 0 < x < 1.

SolutionSuppose that f (x) = 1/x is uniformly continuous in R, then for all ε > 0, there would exist a δ > 0 such that | 1/x1

– 1/x2| < ε whenever |x1-x2| < δ where x1, x2 ϵ (0, 1). ε > | 1/x1

– 1/x2| = (|x2-x1| )/ (x1x2)Or |x1-x2| < x1x2 εTherefore, to satisfy the definition of uniform continuity we would have to have δ ≤ x1x2 ε for all x1, x2 ϵ (0, 1), but that would mean that δ ≤ 0. Therefore there is no single δ > 0. Therefore f (x) = 1/x is not uniformly continuous in 0 < x < 1.

Lipschitz continuity

138


139


140


141

since


142

since

143

Thank you allThank you all

Title: Functions, Limits and Continuity Prof. Dr. Nasima Akhter And Md. Masum Murshed Lecturer Department of Mathematics, R.U. 29 July, 2011, Friday Time:

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