LimitContinuity.mws

Continuity

>    restart;
with( plots ):

Warning, the name changecoords has been redefined

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Lesson Overview

Limits are no longer an unfamiliar concept for us. We understand them conceptually, rigorously, and graphically. We have a substantial collection of Limit Laws and techniques for manipulating and evaluating limits. In this lesson the limit will be used to define continuity at a point and on an interval. These concepts are not difficult to learn and will be used repeatedly throughout the remainder of this course --- and much of higher mathematics.

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Continuity at a Point

Definition - Continuity at a Point

A function f is continuous at a point   a  if the following three conditions are satisfied:

   i) a  is in the domain of f, i.e., f(a)  exists

  ii) limit(f(x),x = a)  exists

 iii) limit(f(x),x = a) = f(a)  

If any one of these three conditions is not satisfied, the function is said to be discontinuous at   a .

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When a function, f, is undefined at an isolated point, a , then a  is called a singularity  of the function. If it is possible to define f(a)  so that the modified function is continous at a  then a  is called a removable singularity of  f; if not, then a  is called an essential singularity  of f.

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Example 1 - Removable Singularity

Consider the function f(x) = sin(x)/x . Because this function is not defined at x = 0  it is discontinuous at 0.

But, because Limit(f(x),x = 0) = 1 , if f(0) = 1 , then the resulting function, namely,

  f(x) = PIECEWISE([sin(x)/x, x <> 0],[1, x = 0])  

is continuous at x = 0 .

A plot of the modified function clearly shows how this definition of f(0)  agrees with the limit at 0.

>    f := x -> piecewise( x=0, 1, sin(x)/x ):
p1 := plot( f, -3*Pi..-0.1 ):
p2 := plot( f, 0.1..3*Pi ):
p3 := pointplot( [0,f(0)], style=point,
                 symbol=circle, symbolsize=18, color=red ):
display( [p1,p2,p3],
         title="Plot showing continuity of modified function at x=0" );

[Maple Plot]

Observe that if the red circle were located anywhere else along the vertical axis the function would be discontinuous at x = 0 .

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Example 2 - Essential Singularity

The function g(x) = sin(x)/abs(x)  is related to the function f in Example 1: g(x) = f(x)  for x  > 0 and g(x) = -f(x)  for x  < 0. Because g(0) is not defined, g is discontinuous at x = 0  because of the singularity at 0.

However, unlike Example 1, it is not possible to define g(0)  to remove the singularity. Another way to show that g is discontinuous at x = 0  is:

limit(g(x),x = 0,right) = 1  and limit(g(x),x = 0,left) = -1  means that limit(g(x),x = 0)  does not exist.

The different values for the one-sided limits means it is not possible to define g(0)  to remove the singularity at 0.

The nature of the essential singularity of this function at 0 is easily seen in the following plot.

>    g := x -> piecewise( x=0, undefined, sin(x)/abs(x) ):
p4 := plot( g, -3*Pi..-0.1 ):
p5 := plot( g, 0.1..3*Pi ):
display( [p4,p5],
         title="Plot showing essential singularity of sin(x)/abs(x) at x=0" );

[Maple Plot]

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If a  is an endpoint of the domain of the function, the limits in conditions ii) and iii) are replaced with the appropriate one-sided limits.

Example 3 - Continuity at Endpoints

Consider the function F(x) = sqrt(-x^2+10*x-21) . The domain of this function is the set of all numbers x  such that -x^2+10*x-21 = (x-3)*(7-x)  >= 0, that is 3 <= x  <= 7.

>    F := x -> sqrt( -x^2+10*x-21 ):
p6 := plot( F, 3.0001..6.9999 ):
p7 := pointplot( [[3,F(3)],[7,F(7)]], symbol=circle, symbolsize=18, color=red ):
display( [p6,p7], view=[2..8,-0.1..2], title="Plot showing continuity at the endpoints of a domain" );

[Maple Plot]

The two-sided limits limit(F(x),x = 3)  and limit(F(x),x = 7)  do not exist. However, because x = 3  is an endpoint of the domain of F, F(3) = 0  and limit(F(x),x = 3,right) = 0  are enough to conclude that F is (right) continuous at x = 3 . Similarly, F is (left) continuous at x = 7  because F(7) = 0  and limit(F(x),x = 7,left) = 0 .

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Example 4 - Continuity at Endpoints II

The function G(x) = 1/sqrt(x-1)  has domain ( 1, infinity  ). Because this domain is open there are no endpoints to consider. Thus, because condition i) is not satisfied, G is not continuous at x = 1 . Moreover, because limit(G(x),x = 1,right) = infinity , there is no possibility to extend the domain to [ 1, infinity  ) to make G continuous at x = 1 .

This example, and the following plot, illustrate that a function cannot be continuous at a point where the function has a vertical asymptote.

>    G := 1/sqrt(x-1):
p8 := plot( G, x=1.01..50 ):
p9 := implicitplot( x=1, x=0..50, y=0..10, color=cyan ):
display( [p8,p9], view=[0..50,0..10],
         title="Plot of function and vertical asymptote; not continuous at x=1" );

[Maple Plot]

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It is tempting to remember only condition iii). That is fine, provided you remember that if either the limit or the function value does not exist then the function is discontinuous at that point.

Continuity on an Interval

Definition - Continuity on an Interval

Let I  be an interval, either closed, open, or half-open, and f a function. We say f is continuous*on*the*interval   I  when f is continuous at every point in I .

Recall that testing for continuity at an endpoint is done using the appropriate one-sided limit instead of the two-sided limit used at all interior points of I.

Geometrically, a function is continuous on an interval if is possible to graph the function in one piece. There can be corners or cusps in the graph but there can be no holes, jumps, or vertical asymptotes. But, as we have seen previously, graphs can be quite misleading.

Because there are an infinite number of points in every interval it is not practical to apply the definition. Sometimes it is not practical to graph a function. Even when it is possible, we have seen examples where it is difficult to obtain an accurate graph of a function.

So, how does one determine an interval where the function is continuous?

Continuity of a function on an interval is typically determined by showing how the function is assembled in appropriate ways from known continuous functions. Because continuity is defined in terms of limits, the building blocks for continuous functions are essentially the same as the building blocks for limits. These results can be summarized in a table that resembles the table of Limit Laws. In the following table, which closely resembles the table of Limit Laws, assume c  is a constant, n  is a positive integer, f is a continuous function on an interval I[1] , and g is a continuous function on an interval I[2] .

 

  MATRIX([
[Name, Formula, `Interval of Continuity`], [___________________, _______________
________, ____________________________], [Constant, F(x) = c, `all real numbers`
 = (-infinity, infinity)], [Ident...  

 

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Example 5

Consider the function

  F(x) = tan((abs(4-x^(3/4))+4*x^2-2*x+3)/(x^2+1))  .

To create a plot of the function showing all asymptotes it helps to know exactly where this function is defined and where it is continuous. We consider the function as the composition F(x) = f(g(x))  where f(x) = tan(x)  and g(x) = (abs(4-x^(3/4))+4*x^2-2*x+3)/(x^2+1) .

>    f := x -> tan( x ):
g := x -> (abs( 4-x^(3/4) ) + 4*x^2-2*x+3) / (x^2+1):
F := unapply( f( g(x) ), x ):

First, identify the domain of the function g. Because the denominator is always positive, there are no problems with the division by zero. The numerator is defined, and continuous, for all x  >=  0.

>    plot( g(x), x=0..20, view=[0..20,0..10], title="Plot of g(x) on its domain" );

[Maple Plot]

To determine if, and where, F has any singularities it is necessary to know if the range of g contains any odd multiples of Pi/2 . To find the range of g, observe that

>    `g(0)` = g(0);

`g(0)` = 7

and

>    q1 := Limit( g(x), x=infinity ):
q1 = value( q1 );

Limit((abs(-4+x^(3/4))+4*x^2-2*x+3)/(x^2+1),x = infinity) = 4

The minimum value of g occurs at

>    q2 := diff( g(x), x ):
q3 := fsolve( q2=0, x ):
x = q3;

x = 2.361158435

with minimum value

>    g(q3);

3.448356739

Thus, the range of g is the closed interval [ 3.448357, 7 ].

Note that the graph of g has a horizontal asymptote of y = 4 . This means y = tan(4)  is a horizontal asymptote for the graph of F.

>    HA := y = limit( F(x), x=infinity ):
HA;

y = tan(4)

The only odd multiple of Pi/2  in the range of g is 3*Pi/2 = 4.712389  and g attains this value only once, near

>    q4 := fsolve( g(x) = 3*Pi/2, x ):
VA := x = q4:
VA;

x = .6400732558

This is a vertical asymptote for F.

 In conclusion, F is continuous on ( 0, 0.640073 ) `union`(``,``) ( 0.640073, infinity  ) = ( 0, infinity  ) `minus`(``,{6.40073}) .

>    p10 := plot( F(x), x=0..q4-0.001, y=-10..10 ):
p11 := plot( F(x), x=q4+0.001..20, y=-10..10 ):
display( [p10,p11], title="Plot of F(x) on its domain" );

[Maple Plot]

We have already seen that the singularity in the domain of F is a vertical asymptote and identified a horizontal asymptote for F. Our final plot includes these asymptotes

>    p12 := implicitplot( {VA,HA}, x=0..20, y=-10..10, color=cyan ):
display( [p10,p11,p12], title="Plot of F and its asymptotes" );

[Maple Plot]

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The table of rules for constructing continuous functions assumes that you start with continuous functions f and g. Some combinations of discontinuous functions can create a continuous function.

Example 6

The functions f and g defined by

  f(x) = PIECEWISE([x-1, x < 1],[5, 1 <= x])  

  g(x) = PIECEWISE([2*x, x < 1],[2-5*x, 1 <= x])  

are discontinuous at x = 1  but their sum,

  f(x)+g(x) = PIECEWISE([3*x-1, x < 1],[7-5*x, 1 <= x])  ,

is continuous for all real numbers.

>    f := piecewise( x<1, x-1, 5 ):
g := piecewise( x<1, 2*x, 2-5*x ):
plot( [f,g,f+g], x=-2..3, discont=true, color=[blue,green,red],
      legend=["y=f(x)","y=g(x)","y=f(x)+g(x)"],
      title="Sum of discontinuous functions can be continuous" );

[Maple Plot]

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Lesson Summary

Continuous functions are generally easy to recognize from a well-drawn plot of the function on an appropriate domain. The problem is that creating a good plot can be difficult without knowing the domain of the function, where it is continuous (and discontinuous), any asymptotes, and other information about the function. Until we develop -- in the Global Analysis and Local Analysis lessons in Unit 3 -- the ability to determine where a function has local and global maxima and minima, is increasing, decreasing, concave up, and concave down, continuity will be determined by using continuity properties of simple functions and the rules for combining continuous functions to create new continuous functions.

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What's Next?

The online practice problems for this lesson provide an opportunity for you to develop your skills and understanding of continuity. The online homework assignment and homework problems from the textbook provide an assessment of your mastery of this topic.

The Intermediate Value Theorem is an ``application'' of continuity that will be used in some of the later lessons in this course.

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