1. Chapter 5
Limits and Continuity

2. Section 5.2
Continuous Functions

3. Definition 5.2.1
Let f : D  and let c  D. We say that f is continuous at c if
for each  > 0 there exists a  > 0 such that | f (x) – f (c)| < 
whenever | x – c | <  and x  D .
If f is continuous at each point of a subset S of D, then f is said to be continuous on S.
Notice that the definition of continuity at a point c requires c to be in D, but it
does not require c to be an accumulation point of D.
In fact, if c is an isolated point of D, then f is automatically continuous at c.
For if c is an isolated point of D, then there exists a  > 0 such that, if | x – c | < 
and x  D, then x = c.
Thus whenever | x – c | <  and x  D, we have
| f (x) – f (c) | = 0 <  for all  > 0. Hence f is continuous at c.

4. Theorem 5.2.2
Let f : D  and let c  D. The following three conditions are equivalent:
(a) f is continuous at c.
(b) If (xn) is any sequence in D such that (xn) converges to c, then
lim n   f (xn) = f (c).
(c) For every neighborhood V of f (c) there exists a neighborhood U of c
such that f (U  D)  V.
And, if c is an accumulation point of D, then the above are all equivalent to
(d) f has a limit at c and lim x  c f (x) = f (c).
Proof:
Suppose first that c is an isolated point of D.
Then there exists a neighborhood U of c such that U  D = {c}.
It follows that,
for any neighborhood V of f (c), f (U  D) = { f (c)}  V. Thus (c) always holds.
Similarly, by Exercise , if (xn) is a sequence in D converging to c, then
xn  U for all n greater than some M.
But this implies that xn = c for n > M,
so limn   f (xn) = f (c). Thus (b) also holds.
We have already observed that (a) applies, so (a), (b), and (c) are all equivalent.

5. Theorem 5.2.2
Let f : D  and let c  D. The following three conditions are equivalent:
(a) f is continuous at c.
(b) If (xn) is any sequence in D such that (xn) converges to c, then
lim n   f (xn) = f (c).
(c) For every neighborhood V of f (c) there exists a neighborhood U of c
such that f (U  D)  V.
And, if c is an accumulation point of D, then the above are all equivalent to
(d) f has a limit at c and lim x  c f (x) = f (c).
Proof:
Now suppose that c is an accumulation point of D.
Then (a)  (d) is Definition 5.1.1,
(d)  (c) is Theorem 5.1.2,
and (d)  (b) is essentially Theorem 
Let’s look at this graphically.

6. f is continuous at c iff
for each  > 0
there exists a  > 0
such that | f (x) – f (c)| <  whenever | x – c | <  and x  D.
Or, in terms of neighborhoods, for each neighborhood V of f (c),
there exists a neighborhood U of c,
such that f (U  D)  V. f
f (c) +  V
f (c)
f (c) –  c
c – 
c +  U

7. Example 5.2.3 Example 5.2.5 It appears that f may be
Let p be a polynomial. We saw in Example that for any c  , lim x  c p(x) = p(c).
It follows that p is continuous on .
Example 5.2.5
Let f (x) = x sin (1/x) for x  0 and f (0) = 0. Here is the graph:
It appears that f may be
continuous at x = 0.
Let’s prove that it is.

8. We claim that the function defined by f (x) = x sin (1/x) for x  0 and f (0) = 0
is continuous at x = 0.
Since for all x,
given  > 0, we may let  =  .
Then when | x – 0 | <  we have | f (x) – f (0)|  | x | <  =  .
The negation of (a) and (b) in Theorem gives a useful test of discontinuity.
Theorem 5.2.6
Let f : D  and let c  D. Then f is discontinuous at c iff there exists a
sequence (xn) in D such that (xn) converges to c but the sequence ( f (xn)) does
not converge to f (c).

9. Example 5.2.8 describes a function (the Dirichlet function) that is discontinuous
at every real number.
1, if x is rational,
0, if x is irrational.
Define f :  by f (x) =
If c  , then every neighborhood of c will
contain rational points at which f (x) = 1 and
also irrational points at which f (x) = 0.
Thus lim x  c f (x) cannot possibly exist and f is discontinuous at each c  .

10. Example 5.2.9 describes a modified Dirichlet function.
Define f : (0, 1)  by f (x) = 1 2 3
f (1/2) = 1/2 
f (1/3) = f (2/3) = 1/3    
f (1/4) = f (3/4) = 1/4        
We claim that f is continuous at each irrational number in (0, 1) and
discontinuous at each rational in (0, 1).

11. c Example 5.2.9 describes a modified Dirichlet function.
Define f : (0, 1)  by f (x) =  1 2 3
Let c be a rational number in (0, 1).
Let (xn) be a sequence of irrationals in
(0, 1) with xn  c.
Then f (xn) = 0 for all n, so
lim f (xn) = 0. c
But f (c) > 0 since c is rational.
So lim f (xn)  f (c) and f is discontinuous at c.

12. 1/k Example 5.2.9 describes a modified Dirichlet function.
Define f : (0, 1)  by f (x) = 1 2 3 
Let d be an irrational number in (0, 1).
Given any  > 0, there exists k 
such that 1/k <  .        
1/k  
There are only a finite number of
rationals in (0, 1) with denominators
less than k.   d
Thus there exists a  > 0 such that all the rationals in (d – , d +  ) have a
denominator (in lowest terms) greater than or equal to k.
It follows that if x  (0, 1) and | x – d | <  , then | f (x) – f (d)| = | f (x)|  1/k <  .
Hence f is continuous at d.

13. Theorem
Let f and g be functions form D to and let c  D. Suppose that f and g
are continuous at c. Then
(a) f + g and f g are continuous at c, and
(b) f /g is continuous at c if g(c)  0.
Proof:
To show that f + g is continuous at c
it suffices by Theorem to show that lim ( f + g)(xn) = ( f + g)(c).
Let (xn) be a sequence in D converging to c.
From Definition and Theorem 4.2.1, we have
lim ( f + g)(xn) = lim [ f (xn) + g (xn)]
= lim  f (xn) + lim g (xn)
=  f (c) + g(c) = ( f + g)(c).
The proofs for the product and quotient are similar, the only difference being that
for f /g we have to choose the sequence (xn) so that g (xn)  0 for all n. Recall that the
quotient is defined only at those points x  D for which g (x)  0. 

14. Theorem 5.2.12 g(V  E)  W. f (U  D)  V.
Let f : D  and g: E  be functions such that f (D)  E. If f is
continuous at a point c  D and g is continuous at f (c), then the composition
g  f : D  is continuous at c.
Proof:
Let W be any neighborhood of
( g  f )(c) = g( f (c)). D E W
Since g is continuous at f (c), there exists
a neighborhood V of f (c) such that
g(V  E)  W. U V
g( f (c)) c g f
f (c)
Since f is continuous at c, there exists
a neighborhood U of c such that
f (U  D)  V.
Since f (D)  E, we have f (U  D)  (V  E).
If follows that g( f (U  D))  W,
so g  f is continuous at c by Theorem

15. Theorem
A function f : D  is continuous on D iff for every open set G in there exists
an open set H in such that H  D = f – 1(G).
Corollary
A function f :  is continuous iff f – 1(G) is open in whenever G is open in . f
f – 1(G） G U V c
f (c)
Proof:
Suppose f is continuous. Let G be an open set in .
We want to show f – 1(G) is open.
Suppose c  f – 1 (G).
Then f (c)  G.
Since G is open, there exists a neighborhood V of f (c) such that V  G.
Since f is continuous, there exists a neighborhood U of c such that f (U)  V  G.
But then U  f – 1(G) so c is an interior point of f – 1(G) and f – 1(G) is open.

16. Theorem
A function f : D  is continuous on D iff for every open set G in there exists
an open set H in such that H  D = f – 1(G).
Corollary
A function f :  is continuous iff f – 1(G) is open in whenever G is open in . f
f – 1(V） U
f (U) V c
f (c)
Conversely, suppose G being open implies f – 1(G) is open.
Given c  , let V be a neighborhood of f (c).
Then V is open.
So, f – 1(V) is open.
Since f (c)  V, c  f – 1(V).
Thus there exists a neighborhood U of c such that U  f – 1(V).
But then f (U )  V , and f is continuous at c. 

17. The following example shows how a discontinuous function might fail to have
the pre-image of an open set be open.
Example
Define f :  by f (x) =
x, if x  2,
4, if x > 2. 4 3 2 1
Let G be the open interval (1, 3).
( )
Then f – 1(G) = (1, 2], which is not open. G
( ]
f – 1(G)