Continuity

　

A function f is continuous at a if

limit(f(x),x = a) = f(a)

Intuitively speaking, f is continuous at a if f(x) approaches f(a) as x approaches a , as shown in the animation below :

　

　

[Maple Plot]

 

　

If f is defined near a , we say f is discontinuous at x = a (or f has a discontinuity at a ), if f is not continuous at a . How can f be discontinuous at a ?

　

Case (i) limit(f(x),x = a) exists but limit(f(x),x = a) <> f(a) or f is not defined at a ,for example f(x) = (x^3-1)/(x-1) at x = 1 .

　

[Maple Plot]

In this case, we say f has a removable discontinuity at x = a , because we can redefine f by f(a) = limit(f(x),x = a) so that f is continuous at x = a .

　

Case (ii) Both limit(f(x),x = a,left) and limit(f(x),x = a,right) exist, but limit(f(x),x = a,left) <> limit(f(x),x = a,right) , for example

　

f(x) = x if x < 1 and f(x) = x+1 if 1 <= x

　

　

[Maple Plot]

In this case, f has a jump discontinuity at x = a .

　

Case (iii) Either limit(f(x),x = a,left) or limit(f(x),x = a,right) does not exist, for example f(x) = sin(1/x) .

　

　

　

[Maple Plot]

Theorem If f and g are continuous at a and c is a constant, then the following functions are also continuous at a :

1. f+g 2. c*f 3. f*g 4. f/g if g(a) <> 0 .

　

Note that from 1 and 2, we get that f-g is also continuous at a .

　

Since limit(P(x)/Q(x),x = a) = P(a)/Q(a) for P , Q polynomials and Q(a) <> 0 , any rational functions is continuous wherever it is defined.

　

We say f is continuous in an interval, if f is continuous at very number in this interval. Geometrically, we can think of a function that is continuous in an interval as a function whose graph has no break in it. In other words, the graph can be drawn without removing your pen from the paper.

　

Intermediate Value Theorem

Suppose that f is continuous on the interval [a, b] and let d be any number between f(a) and f(b) , where f(a) <> f(b) . Then there exists a number c in ( a, b ) such that f(c) = d .

　

　

　

[Maple Plot]

 

　

The Intermediate Value Theorem says that if d[1], d[2] , where d[1] < d[2] , are in the range of a continuous function f , then the interval [d[1], d[2]] is contained in the range of f .

　

[Maple Plot]

 

　

We can use the Intermediate Value Theorem to locate roots of equations as in the following example :

Consider the equation 4*x^3-6*x^2+3*x-2 = 0 , since

　

f(1) = 4-6+3-2*1 = -1 and f(2) = 32-24+6-2 = 12

　

The Intermediate Value Theorem says that there is a number c in ( 1, 2 ) such that f(c) = 0 . In other words, the equation has at least one root in the interval ( 1, 2 ). In fact, the Intermediate Value Theorem can help us finding the numerical values of these roots using bisecting algorithm .

1.221122742, .7629394531e-5, 17

　

The Inverse of a Continuous Function

　

From the animation below, it is not hard to believe that the inverse of a continuous function is also continuous.

　

 

[Maple Plot]

　

Suppose that f is strictly increasing and continuous on [a, b] . Let c = f(a) and d = f(b) and let g be the inverse of f . Note that g is strictly increasing on [c, d] . Let y[0] be a number in ( c, d ), to show g is continuous at y[0] , we have to show that for each epsilon > 0 there is a delta > 0 such that

　

g(y[0])-epsilon < g(y) < g(y[0])+epsilon whenever y[0]-delta < y < y[0]+delta

　

Let x[0] = g(y[0]) , so that f(x[0]) = y[0] . Suppose epsilon is given, without loss of generality, we may assume that epsilon is very small that both x[0]-epsilon and x[0]+epsilon are in [a, b] .

　

Let delta = min(f(x[0])-f(x[0]-epsilon),f(x[0]+epsilon)... , see the diagram below .

[Maple OLE 2.0 Object]

　

If y[0]-delta < y < y[0]+delta then f(x[0]-epsilon) < y < f(x[0]+epsilon) . Since is strictly increasing, we get

　

g(y[0])-epsilon = x[0]-epsilon = g(f(x[0]-epsilon)) < g(y) < g(f(x[0]+epsilon)) = x[0]+epsilon = g(y[0])+epsilon

　

Therefore, g is continuous at y[0] .

For each integer n , since f(x) = x^n is continuous and increasing on [ 0, infinity ), its inverse function g(x) = x^(1/n) is also continuous on [ 0, infinity ).

　

　

Question : Suppose that f is a one-to-one continuous function defined on [a, b] , is it true that f is monotonic on [a, b] (that is, f is increasing or f is decreasing on [a, b] ) ?

　

The Intermediate Value Theorem can help us answer this question. Suppose that f is not monotonic on [a, b] , then there exist c[1], c[2], c[3] in [a, b] with c[1] < c[2] < c[3] such that

　

f(c[1]) < f(c[2]) and f(c[3]) < f(c[2])

or

f(c[2]) < f(c[1]) and f(c[2]) < f(c[3])

　

Without loss of generality, we may assume f(c[1]) < f(c[2]) and f(c[3]) < f(c[2]) . Let d be a number between max(f(c[1]),f(c[3])) and f(c[2]) , the Intermediate Value Theorem says that there exist numbers c[4], c[5] with c[4] in ( c[1], c[2] ), c[5] in ( c[2], c[3] ) such that f(c[4]) = f(c[5]) = d , as shown in the graph below, which contradicts to the fact that f is one-to-one. Hence, f is monotonic on [a, b] .

　

　

　

[Maple Plot]

 

　

Assume that limit(g(x),x = a) = b and f is continuous at b .

Given epsilon > 0, since f is continuous at b , there exists delta[1] > 0 such that

　

abs(f(y)-f(b)) < epsilon whenever abs(y-b) < delta[1]

　

Since limit(g(x),x = a) = b , there exists delta > 0 such that

abs(g(x)-b) < delta[1] whenever 0 < abs(x-a) < delta

　

[Maple OLE 2.0 Object]

　

Hence, if 0 < abs(x-a) < delta , we have abs(g(x)-b) < delta[1] and then abs(f(g(x))-f(b)) < epsilon . So, we have the following :

Theorem If f is continuous at b and limit(g(x),x = a) = b , then limit(f(g(x)),x = a) = f(b) . That is,

　

limit(f(g(x)),x = a) = f(limit(g(x),x = a))

　

In other words, the limit symbol can be move through the function symbol.

　

Consequently, If f is continuous at b and g is continuous at f(b) , then the composition function f o g is continuous at a .

　

　

Example Show that the function h(x) = sqrt(1-x^2) is continuous on [ -1, 1 ].

　

Solution :

Since g(x) = 1-x^2 is continuous on on [ -1, 1 ] , 0 <= g(x) for all x in on [ -1, 1 ] and f(x) = sqrt(x) is continuous for all 0 <= x . By the above theorem, h(x) = f(g(x)) is continuous on [ -1, 1 ].

　

　

Question : Suppose that limit(g(x),x = a) = b , f is defined at b and limit(f(x),x = b) exists, is it true that

　

limit(f(g(x)),x = a) = f(limit(g(x),x = a)) ?

　

　

The following example show that if we replace the continuity condition on f by the existence of limit(f(x),x = b) , then the above theorem will not be true even that f is well defined at b .

If f(x) = PIECEWISE([x^2, x <> 0],[4, x = 0]) and g(x) = 2*x , then f(g(x)) = PIECEWISE([4*x^2, x <> 0],[4, x = 0]) . We get

　

limit(f(g(x)),x = 0) = 0 and f(limit(g(x),x = 0)) = 4 .

　

Therefore, limit(f(g(x)),x = 0) <> f(limit(g(x),x = 0)) .

　

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