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The concept of

continuityRT1 - St. Arnold Janssen

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Definitions

According to Euler -A continuous curve is one such that its naturecan be expressed by a single function ofx. If a curve is of such a naturethat for its various parts ... different functions ofxare required for itsexpression, ... , then we call such a curve discontinuous.

According to Bolzano - A function f(x) varies according to the lawof continuity for all values ofxinside or outside certain limits ... if [when]xis some such value, the difference f(x+ omegalc ) - f(x) can be madesmaller than any given quantity provided omegalc can be taken as small aswe please.

According to Cauchy-The function fwill be, between two assignedvalues of the variablex, a continuous function if for each value ofxbetween these limits the [absolute] value of the difference f(x+ a ) - f(x)decreases indefinitely with a .

According to Dirichlet - One thinks ofa and b as two fixedvalues and ofxas a variable quantity that can progressively take allvalues lying between a and b. Now if to everyxthere corresponds a

single, finite y in such a way that, asxcontinuously passes throughthe interval from a to b,y= f(x) also gradually changes, thenyis

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The Concept ofContinuity

The goal of this section is to introduce theconcept of continuity.

Graphically, a function is said to be continuous if

its graph has no holes, jumps, orincreases/decreases without bound at a certainpoint.

Stated differently, a continuous function has a

graph which can be drawn without lifting thepencil from the paper.

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Continuity at a Point

We say that a function f(x) is continuous atx = a if and onlyif the functional values f(x) get closer and closer to thevalue f(a) as x is sufficiently close to a. We write:

limx a (x) = f(a).

This means, that for any given > 0 we can find a > 0 such that:

|x-a| < implies |f(x)-f(a)| < .

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In words, we say fis continuous at a" if, for eachopen intervalJ containing f(a), we can find an open

interval I containing a so that for each pointxin I, f(x)lies in the intervalJ.

See Figure 8.1

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Example 8.1

(i) Property 5 of the previous section shows thatpolynomials are continuous everywhere.

(ii) Similarly, ifh(x) = f(x)/g(x) is a rational function and a isa number such that g(a) 0 then

limx a f(x)/g(x) = f(a)/g(a)

That is, h(x) is continuous at numbers that are not zeros ofthe function g(x).

(iii) Trigonometric functions are continuous at values wherethey are defined.

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Discontinuity:

A function f(x) that is not continuous atx = a is saidto be discontinuous there. We exhibit threeexamples of discontinuous functions.

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Example 8.2 (RemovableDiscontinuity)

Show that the function f(x) = x2+x-2/x-1 isdiscontinuous atx= 1.

SOLUTION:

Graphing the given function (see Figure 8.2) we find

Fi ure 8.2

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The small circle indicates an excluded point on thegraph. Thus, we see that f(1) is undefined andtherefore f(x) is discontinuous atx = 1. Note that

limx 1f(x) = 3

Thus, if we redefine f(x) in such a way that f(1) = 3then we create a continuous function atx = 1. Thatis, the discontinuity is removable

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Example 8.3 (InfiniteDiscontinuity)

Show that f(x) = 1/xis discontinuous atx = 0.

Solution:

According to Figure 7.4, we have that limx 0 1 /x does notexist. Thus, f(x) is discontinuous atx = 0. Since limx 0 f(x)= , we callx = 0 an infinite discontinuity.

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Example 8.4 (JumpDiscontinuity)

Show that f(x) = |x|/xdiscontinuous atx = 0.

SOLUTION:

The fact that f(x) is discontinuous atx= 0 follows fromFigure 8.3 below

Figure 8.3

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Theorem 8.1

Iffand g are two continuous functions atx = a and kis aconstant then all of the following functions are continuousatx = a.

Scalar Multiple kf

Sum and Difference f g

Productf *g

Quotientf /g , provided that g(a) 0.

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Theorem 8.2 (Continuity ofComposite Functions)

The composition of two continuous functions is continuous.

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Determine iff(x) = 3x2 2 iscontinuous atx= -1

The three conditions must be true:

1. The limit of the function asxapproaches -1 is

limx -1 f(x) = lim x -1 (3 x2 2)

= lim x -1 3 x2 - lim x -1 2

= 3(-1)2 2 = 1

2. The value of the function atx= -1 is

f(-1) = 3(-1)2 2

= 3 2

= 1

3. Since limx -1 f(x) = f(-1) = 1, then the function f(x) =

3x2 2 is

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Continuity on an Interval

We say that a function fis continuous on the openinterval (a,b) if it is continuous at each number in thisinterval.

If in addition, the function is continuous from the right of a,

i.e. limx a

+ f(x) = f(a), then we say that fis continuous onthe interval [a, b).

Iffis continuous from the left ofb,i.e. limx b- f(x) = f(b)then we say that fis continuous on the interval (a,b].

Finally, iff

is continuous on the open interval (a, b

), fromthe right at a and from the left at b then we say that fiscontinuous in the interval [a, b].

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Example 8.5

Find the interval(s) on which each of the given functions iscontinuous:

(i) f1(x) =x2-1/x2-4.

Solution: (-,-2) U (-2,2) U (2, )

(ii) f2(x) = csc x

Solution:x n where n is an integer.

(iii) f3(x) = sin ( 1/x ).

Solution: (- ,0) U (0,)

(iv) f5(x) ={3-x, if -5 x < 2 x-2, if 2 x < 5

Solution: Since limx 2- f5(x) = limx 2- (3-x) = 1 andlimx 2+ (x-2) = 0, f5 is continuous on the interval [-5,2) U

(2,5)

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The Intermediate ValueTheorem

Continuity can be a very useful tool in solving equations.

So if a function is continuous on an interval and changessign then definitely it has to cross the x-axis.

This shows that the function possesses a zero in thatinterval.

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Theorem 8.3 (IntermediateValue Theorem)

Let fbe a continuous function on [a,b] with f(a) < f(b): Iff(a)< d < f(b) then there is a c b such that f(c) = d.

Example 8.6

Show that cosx=x3 -xhas at least one zero on theinterval [/4 , /2].

Solution

Let f(x) = cosx- x3 +x. Since f(/4) 1.008 > 0 and f(/2) -2.305 < 0, by the IVT there is at least one number c in the

interval (/4, /2 ) such that f(c) = 0

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EXERCISES

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Determine if the polynomial or rational function iscontinuous at the given values ofc. If the function is

continuous, find the limit asxapproaches c.

1. f(x) = 2 5x3; c = 1

2. f(x) =x2 3x 1; c = 1

3. f(x) =x3 2x2 + 3x 4; c = 2

4. f(x) = 2x5 + 3x4 8x3 +x2 + 5x 9; c = -1

5. f(x) =x7 + 2x6 4x5 +x4 5x3 +x2 3x+ 8; c = -1

6. f(x) = 3x/x+2; c = 1

7. f(x) =x2 9/x 3; c = 2

8. f(x) =x2 + 6x+ 5/x 3; c = -2

9. f(x) =x/x+ 1; c = 0

10. f(x) =x2/x2 2; c = 1