HIGH SCHOOL DIFFERENTIAL CALCULUS COURSE
3.4. CONTINUITY OF A FUNCTION 3.1.1 Define the continuity of a function at a point 3.1.2 Determine if a function is continuous or discontinuous at a given point from its algebraic expression or its graph Let f be a function and c Domain( f ) . We say that f is continuous at x= c if and only if limf ( x )= f ( c ) limf ( x )= lim f ( x ) = L limf ( x ) = L x c . Knowing that if x c- x c + it can be said that x c , so limf ( x )= lim f ( x ) = lim f ( x ) = f ( c ) = L . the condition of a function to be continuous is x c- x c + x c
If the function f is not continuous at x=c, then f is said to have a discontinuity at c. There are two types of discontinuties. - Removable discontinuity - Nonremovable discontinuity o Jump discontinuity o End of domain o Infinite discontinuity
1) Removable discontinuity is the discontinuity at a point which could be removed by defining the function at just that one point. This often appears as a graph with a hole in it(gap). A removable discontinuity it appears when
x2 - 4 An example of a removable discontinuity could be considered the function f( x ) = in the point x=2. x - 2 x2 - 4骣 0 ( x-2)( x + 2) lim=琪 = lim = lim( x + 2) = 4 x 2-x-2桫 0 x 2 - x - 2 x 2 -
x2 - 4骣 0 ( x-2)( x + 2) lim=琪 = lim = lim( x + 2) = 4 x 2+x-2桫 0 x 2 + x - 2 x 2 +
x2 - 4 lim= 4 x 2 x - 2
f(2)=dne
x2-4 x 2 - 4 x 2 - 4 so because lim= lim = lim = 4 and x 2-x-2 x 2 + x - 2x 2 x - 2 f(2)=dne, the function is not continuous in x=2 and it has a removable discontinuity.
x2 - 4 if x 2 f( x ) = x - 2 in x=2 2if x = 2
x2 - 4骣 0 ( x-2)( x + 2) lim=琪 = lim = lim( x + 2) = 4 x 2-x-2桫 0 x 2 - x - 2 x 2 -
x2 - 4骣 0 ( x-2)( x + 2) lim=琪 = lim = lim( x + 2) = 4 x 2+x-2桫 0 x 2 + x - 2 x 2 +
x2 - 4 so lim= 4 x 2 x - 2
f(2)=2
so because
x2-4 x 2 - 4 x 2 - 4 lim= lim = lim = 4 and x 2-x-2 x 2 + x - 2x 2 x - 2 f(2)=2, the function is not continuous in x=2 and it has a removable discontinuity. 2) Nonremovable discontinuity is the discontinuity at a point which could not be removed at just that point. There are three types of nonremovable discontinuities: 2a) Jump discontinuity is usually caused by a piecewise-defined function whose pieces don't really meet together.
limf ( x ) = L1 limf ( x ) = L2 L L In other words x c- and x c+ and 1 2 -x - 2 if x �( ,1) Example1: f( x ) = x- 2 if x [ 1,4] lim(-x - 2) =( - 1 - 2) = - 3 x 1-
limx - 2 = 1 - 2 = - 1 x 1- ( ) ( ) . because the limit from the left is not equal with the limit from the right
limf ( x ) = dne x 1
and f(1)= -1
The function is not continuous in the x=1 and it has a jump discontinuity 2b) End of domain is when the function has a limit from one side, and from the other side does not exist.
In other words y
limf ( x ) = L limf ( x ) = dne x c- and x c+
OR
x
limf ( x ) = dne limf ( x ) = L - and + x c x c
An example of end of domain discontinuity is: f( x )= x - 2 + 1 limx- 2 + 1 = 0- + 1 = dne x 2- ( )
limx - 2 + 1 = 0+ + 1 = 1 x 2+ ( )
so limx- 2 + 1 = dne x 2 ( ) The value of the function in x=2 is f(2)=1
Because the limit does not exist from the left and the limit from the right is 1, the type of discontinuity is end of domain.
2c) Infinite discontinuity is defined where the jump is to the infinity
In other words
limf ( x ) = limf ( x ) = - limf ( x ) = - limf ( x ) = x c- and x c+ OR x c- and x c+ y
x
1 f( x ) = in x=0 x y 骣1 骣 1 lim 琪= 琪 - = - x 0- 桫x 桫0
骣1 骣 1 lim 琪= 琪 + = + x 0+ 桫x 桫0
x 骣1 骣 1 lim琪= 琪 = dne x 0 桫x 桫0
骣1 f(0) =琪 = dne 桫0
Because the limit from the left is not equal with the limit from the right, the function is not continuous in x=0 and the type of discontinuity is infinite discontinuity. Discontinuity at a point.
limf ( x ) f ( c ) A function has a discontinuity at a point x=c if and only if x c Discontinuity on an interval
A function f is continuous on the closed interval [a, b] if it is continuous on the open a, b limf ( x )= f ( a ) limf ( x )= f ( b ) interval ( ) and x a+ and x b+ Example1:
Given the following piecewise function
2x+ 1 if x �( � , 2] f( x )= 1 - x2 if x �( 2,2) x+2 + 1 if x �[ 2, ) a) Analyze the function in each step point and decide the continuity or the type of discontinuity( if it exists) b) Decide if the function is continuous in its entirety c) Graph the function d) Range of the function a) To analyze the function in its step-points, it is needed to check the continuity in these step points (Step-points are the points of a piecewise function, where a component function is finishing, and other component function is starting). In our case, there are two step-points where is needed to check the continuity, at x=-2 and x=2. In x=-2 From the left of x=-2, lim (2x + 1) = 2�( + 2) = 1 - + 4 = 1 - 3 and from the right, x� 2- lim (1-x2 ) = 1 - ( - 2) 2 = 1 - 4 = - 3 + ( ) . Because the limit from the left and the limit x� 2 limf ( x )= - 3 from the right has the same value, it can be decided that x� 2 The value of the function at x=-2 is f (- 2) = 2�( + 2) = 1 - + 4 = 1 - 3. lim (2x+ 1) = f ( - 2) = - 3 Because of x� 2 it can be concluded that the piecewise function is continuous in x=-2
In x=2 2 2 From the left of x=2, lim(1-x ) = 1 - 2 = - 3 and from the right, x 2- ( ) limx + 2 + 1 = 2 + 2 + 1 = 2 + 1 = 3 . Because the limit from the left and the limit x 2+ ( ) ( ) from the right has different values, in this point there is a discontinuity. So, the limf ( x ) = dne limit x 2 . The value of the function at x=2 is f (2)= 2 + 2 + 1 = 4 + 1 = 2 + 1 = 3. limf ( x ) = dne The limit x 2 because of the limit from the left has a different value than the limit from the right and it can be concluded that the piecewise function has a jump discontinuity at x=-2 b) To decide the continuity in its entirety, it is needed to check each function´s continuity. The first function, f( x )= 2 x + 1 for x �( � , 2] is continuous in its entirety of domain, the second function f( x )= 1 - x2 for x �( 2,2) also is continuous in its entirety of domain and the third function f( x )= x + 2 + 1 for x �[ 2, ) is also continuous in its domain. y
d)
The range of the function is:
y �(ゥ ,1]U[ 3, )
x