Continuous Functions
Definition:
Let A , let f : A → , and let c A . We say that f is continuous at c if, given any number
0 there exists 0 such that if x is any point of A satisfying | x − c | , then
| f (x ) − f (c ) | .
If f fails to be continuous at c , then we say that f is discontinuous at c .
Remarks:
1- If c A is a cluster point of A , then a comparison of Definition of limit functions and
Definition of continuous functions, show that f is continuous at c if and only if
lim f (x ) = f (c ) .
x →c
Thus, if c A is a cluster point of A , then three conditions must hold for f to be
continuous at c :
(i) f (c ) makes sense,
(ii) lim f (x ) makes sense, and
x →c
(iii) lim f (x ) = f (c ) .
x →c
2- If c A is isolated point (that is if c A is not cluster point) , then there exist a
neighborhood V (c ) such that A V (c ) = c .
3- If f : A → and c A is isolated point, then f is continuous at c . Indeed, if c A is
isolated point, then there exist 0 such that A (c − , c + ) = c . Then
f (x ) − f (c ) = 0 for all x A (c − , c + ) = c ,
And so lim f (x ) = f (c ) .
x →c
4- Since continuity is automatic for isolated points, we generally test for continuity only at
cluster points.
Theorem:
A function f : A → is continuous at c A if and only if for every sequence x n in A that
converges to c , the sequence f (x n ) converges to f (c ) .
Definition:
If A and let f : A → . If B is subset of A , we say that f is continuous on B if it is
continuous at every point of B .
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Combinations of Continuous Functions
Theorem:
Let A , let f , g : A → , and let k . If f and g are continuous at c A . Then
1- f + g is continuous at c .
2- f − g is continuous at c .
3- kf is continuous at c .
4- f g is continuous at c .
f
5- is continuous at c ; g (c ) 0 .
g
Proof:
If c A is isolated point, then f + g , f − g , kf , fg , f / g ; g (c ) 0 are automatic continuous at c .
Hence, suppose that c A is cluster point. Since f and g are continuous at c A , then
lim f (x ) = f (c ) and lim g (x ) = g (c ) . Then
x →c x →c
1) lim (f + g )(x ) = lim (f (x ) + g (x )) = lim f (x ) + lim g (x ) = f (c ) + g (c ) = (f + g )(c ) , that
x →c x →c x →c x →c
is f + g is continuous at c .
2) lim (f − g )(x ) = lim (f (x ) − g (x )) = lim f (x ) − lim g (x ) = f (c ) − g (c ) = (f − g )(c ) , that
x →c x →c x →c x →c
is f − g is continuous at c .
3) lim (kf )(x ) = lim kf (x ) = k lim f (x ) = kf (c ) = (kf )(c ) , that is kf is continuous at c .
x →c x →c x →c
4) lim (f g )(x ) = lim (f (x ) g (x )) = lim f (x ) lim g (x ) = f (c ) g (c ) = (f g )(c ) , that is
x →c x →c x →c x →c
f g is continuous at c .
f lim f (x )
f (x ) x →c f (c ) f f
5) lim ( x ) = lim = = = (c ) , that is , g (c ) 0 ,is continuous
x →c g x →c g ( x ) lim g (x ) g (c ) g g
x →c
at c .
Theorem:
Let A , let f , g : A → , and let k . If f and g are continuous on A . Then
f
f + g , f − g , kf , fg and , g (x ) 0 for all x A , are continuous on A .
g
Theorem:
Let A , B , let f : A → , g :B → be functions such that f (A ) B . If f is continuous at
a point c A and g is continuous at f (c ) B , then the composition g f : A → is continuous
at c .
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Proof:
Let x n be a sequence in A such that x n → c . Now we need to prove that
( g f )(x n ) → ( g f )(c ) and so g f is continuous at c . Since x n → c and f is continuous at c
, then f (x n ) → f (c ) . Since f (x n ) → f (c ) and g is continuous at f (c ) , then
g (f (x n )) → g (f (c )) , that is ( g f )(x n ) → ( g f )(c ) . Thus g f is continuous at c .
Theorem:
Let A , B , let f : A → , g :B → be functions such that f (A ) B . If f is continuous on
A and g is continuous on B , then the composition g f : A → is continuous on A .
Examples:
1- The function f (x ) =| x | is continuous on . Indeed, since
| x | − | c | | x − c | for all x , c ,
then, if we take = , the function f (x ) =| x | is continuous on .
2- The function f (x ) = x is continuous on [0, ) . Indeed, if x n 0 for all n such that
x n → x , then xn → x .
Example:
If the function f : A → is continuous on A , prove that the function | f | is defined by
| f | (x ) :=| f (x ) | , for all x A , is continuous on A .
Proof:
Let g (x ) =| x | . Then the function g (x ) is continuous on . Since f is continuous on A and
f (A ) , and g (x ) is continuous on , then the function g f =| f | is continuous on A .
Example:
If the function f : A → is continuous on A , prove that the function f is defined by
( f ) (x ) := f (x ) , for all x A , is continuous on A .
Proof:
Let g (x ) = x . Then the function g (x ) is continuous on [0, ) . If f is continuous on A and
f (x ) 0 for x A and g (x ) is continuous on [0, ) , then the function g f = f is continuous
on A .
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