1.3 Continuity, One-Sided Limits, Intermediate Value Theorem
☼Today’s notes will refer back to this graph throughout.
0
-3
2
4
5
A. One-Sided Limits
Find each of the following:
1. f  3  undefined

f  x  1
 2. xlim
3

f  x  1
3. xlim

 3
f  x  1
 4. xlim
3
1-4 Function undefined, limit exists.
13. f  4   2

f  x  2
14. xlim
 4

f  x  2
15. xlim
 4

f  x  2
16.lim
x 4
13-16, function value and limit match
5. f  0   3

f  x   2
6. xlim
 0

f  x  3
7. xlim

 0
f  x   DNE
8. lim
x 0
5-8, Function defined, Limit DNE
9. f  2   1

f  x  3
10. xlim
 2

f  x  3
11. xlim
 2

f  x  3
12.lim
x2
9-12, both function defined and limit exists.
17. f  5  undefined

f  x   DNE
18. xlim
 5

f  x   DNE
19. xlim
 5

f  x   DNE
 20.lim
x 5
17-20, function is undefined and limit does not exist!
B. Continuity
**Examples from graph**
II. Definition of Continuity at a Point:
*A function is continuous at “X=C” if …
a) f(c) is defined
* y=f(x) is not continuous at (-3) or (5)
b) the
c)
lim f  x  exists.
*y=f(0) is discontinuous at x=0 because of break so limit DNE
x c
lim f  x   f  c 
x c
Cannot draw graph of y=f(0) with one continuous line.
Function value and limit value have to match for it to be continuous.
c) really implies the first two.
* A function is continuous on an open interval (a, b) if the function is continuous for all “c” in (a, b).
**Graph Example** f(x) is continuous on (-3, 0)
*A function is continuous on a closed interval [a, b], if the function is continuous on (a, b) AND
and
lim f  x  exists.
lim f  x  exists
x a
x b
**Graph Example** f(x) is continuous on [3, 4] and not continuous on [4, 5]
III. Removable and Non-Removable Discontinuities
*A function has a removable discontinuity at X=C if we can make “f” continuous at “C” by changing the definition
of “f” at only one point. (By moving or inserting one point)
**Graph Example** 4 discontinuous at -3, 0, 2, 5
Two removable and two non-removable
-3--- removable , 0--- non-removable
2--- removable,
5--- non-removable
IV. Some Important Results: Families of functions that are continuous…
1. Polynomials are continuous everywhere, because the
lim f  x   f  c  I.e. direct substitution always works.
x c
2. If ‘f” and ‘g’ are continuous at ‘c’ then a) f  g is continuous at ‘c’, b) f  g is continuous at ‘c’,
c) f  g is continuous at ‘c’, and d)
f
is continuous at ‘c’ if g  c   0 .
g
3. Since a rational function is the ratio of two polynomials then 1. And 2d) imply a rational function is continuous
for all ‘X’ in its domain.
4. If
lim g  x   L and if ‘f’ is continuous at L then lim f  g  x    f  L  .
x c
That is to say
x c


lim f  g  x    f  lim g  x  
x c
 x c 
5. If ‘g’ is continuous at ‘c’ and ‘f’ is continuous at g(c) then f
6. The absolute value of continuous functions is continuous.
g is continuous at ‘c’.
Which functions are continuous?
 x2  2x
1. f  x   
x  4
x  1
x  1
Solution: All polynomials are continuous so each “piece” is continuous on its piece of the domain. What we have to check is
do the two “pieces” match up (same y value) when x  1 .
We will substitute – 1 into both “pieces” and see if they yield the same y value or a different y value.
x 2  2 x  x  4 when
 1
2
x  1?
?
 2  1  1  4
?
1  2  1  4
3  3 YES !
So we now know that this function is continuous everywhere!
x0
cos x
2. g  x    2
2 x  3x  2 x  0
Solution: The first piece is the cosine function which is continuous on its domain and the second piece is a polynomial
function which is continuous on its domain. The question we need to answer is do the two pieces match up when x = 0.
We will substitute 0 into both “pieces” and see if we get the same y-value.
?
cos x  2 x 2  3 x  2 when
?
x0
cos  0   2  0   3  0   2
1  2
2
No !!!!
So this function is NOT continuous at x = 0 (BUT it is continuous at every other x value).
 x2  x

3. h  x    x  1
x  2

x 1
x 1
We note immediately that at x  1 this function has a vertical asymptote so it will be discontinuous at that value of x (note
that the portion of the domain for the rational piece does include – 1 . What we will check next is to see if this function also
has a discontinuity when x = 1 (i.e. do the two pieces match up or not).
x2  x ?
 x  2 when
x 1
x 1
1  1 ?
 1  2
1  1
2
0  3 No !!!
So this function has two places where it is not continuous (at x = 1 and at x = - 1 )
There are two types of discontinuities….. removable and non-removable. The previous two functions have discontinuities
that are non-removable (vertical asymptotes and “jumps / breaks” are non-removable). In order to be removable it must be
the case that we can redefine the function for exactly one value of x and make it continuous.
Examples of functions with removable discontinuities..
x2  4
this function has a removable discontinuity at x = 2.
x2
“How” do you remove it? Just redefine this function so that you only change what happens at x = 2 BUT the new definition
agrees with the original one everywhere else.
1. k  x  
x 2  4  x  2  x  2 

 x  2 x  2 so if we just make our “new” function k  x   x  2 we have a function that is
x2
 x  2
continuous everywhere and agrees with our original function at every x value except when x = 2. This is an example to show
you that a “hole” is a removable discontinuity.
k  x 
 x2  9
x  3

2. L  x    x  3
This function is a line with a hole in it (first piece) and a single point (second piece) that lies
 0
x  3

directly “above” the hole. In order to “remove” this discontinuity we just need to adjust the single point to fill in the hole.
Go find out where the hole is (y-value).
x 2  9  x  3 x  3

 x  3 x  3 The hole here will be at a "height" equal to x - 3 evaluated at -3
x3
 x  3
that is -3-3=-6
 x2  9

So if we wish to “remove” the discontinuity all we need to do is define L  x    x  3
 -6

x  3
OR L  x   x  3
x  3
Some more problems we can use to explore the idea of continuity.
Find any value(s) of “a” that make the function continuous.
2ax  5 x  3
1. f  x   
3x  a x  3
Each “piece” is continuous on its own domain as each piece is a polynomial (linear to be more specific). We need to find the
value of “a” that will force the two lines to match up when x = 3 (the place where the domain is split).
2ax  5  3 x  a when
x3
2a  3  5  3  3  a
6a  5  9  a
4
a
5
 8
 x5 x  3
4
 5
So if a   out function becomes f  x   
and if you substitute “3” into each “piece” you will see you get
5
3 x  4 x  3

5
49
back the same “y” value, namely 
5
a 2 x
x2
2. f  x   
x

a

1
x2

Again, each “piece” is continuous on its own domain as each piece is a polynomial (linear again). We need to find the value
of “a” that will force the two lines to match up when x = 2 (the place where the domain is split).
a 2 x  x  a  1 when
x2
a  2   2  a  1
2
2a 2  a  3
2a 2  a  3  0
 2a  3 a  1  0
a
3
2
a  1
So apparently there are two different values that “a” could be that would make this function to be continuous.
If a  1
3
If a 
2
x
f  x  
x
x2
so clearly this is continuous.
x2
9
 x
f  x   4
x  5

2
x2
this one is not quite so obvious BUT if you substitute x = 2 into both “pieces” you will get
x2
back the same exact “y” value, namely
9
.
2
C. The Intermediate Value Theorem
If f  x  is a continuous function on  a, b and if f  a   d  f  b  , then there exists a c   a, b such that f  c   d .
The green boxed pieces are called the “hypotheses” (plural of “hypothesis”) of the theorem (all of the words between “if” and
“then”). The red boxed piece is called the “conclusion” of the theorem (all of the words after “then”). In order to be
guaranteed the conclusion to ANY theorem you must first have the hypotheses. You will notice that part of the hypotheses of
this particular theorem is that you have a continuous function on a closed interval. This will be a part of the hypotheses of
many of our theorems in calculus (the concept of continuity is very important).
f(b)
Graphically this theorem means…
d
f(a)
a
c
b
Notice that if the “hypotheses” are not satisfied then you are not guaranteed the conclusion!
f(b)
f(b)
d
d
f(a)
f(a)
a
b
a
b
c
Why does each example above NOT satisfy the “hypotheses” of the IVT?
Notice in the left picture there is no “c” period that satisfies the IVT. Why is that?
Notice in the right picture there is a c where f(c)=d BUT the c is not between a and b (thus not satisfying the IVT).
Answers: In the left picture, since the function is NOT continuous on the closed interval we are not guaranteed
the result of the IVT (although there are certainly some “d’s” that would have a “c” in the interval from “a” to “b”).
In the right picture the graph is NOT a function!
Example: Use the Intermediate Value Theorem to show that f  x   x2  2x  3 has a zero in 0, 2
Solution: 1. Note that the function is continuous as it is a polynomial
2. Note that f(0) = - 3 and that f(2) = 5
3. Note that 0 is between -3 and 5.
So we have a continuous function on a closed interval where f(a) is negative and f(b) is positive so by the IVT
there MUST be at least one c value in the interval where f(c) = 0.