ACOW LIMITS AND CONTINUITY MODULE
Updated 5/28/2016
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Activity C1 – Continuity of a Function Given the Graph
Intuitively, a function is continuous if its graph does not have any holes, breaks, or gaps.
If the graph of a function does have holes, breaks, or gaps, we say the function is
discontinuous. The graph(RS1), for example, is discontinuous at x  2 and x  2
because there is a gap in the graph at each of these values. Using the definition of
continuity(RS2) we would say the previous graph(RS3) is discontinuous at x  2 and
x  2 because lim f ( x) and lim f ( x ) do not exist.
x 2
x2
We can use the limit applet to help us determine if a function is continuous by finding
lim f ( x ) and f ( a ) . If both lim f ( x ) and f ( a ) are real numbers such that
xa
xa
lim f ( x ) = f ( a ) , the function is continuous at x = a. If lim f ( x ) ≠ f ( a ) , the function is
xa
xa
discontinuous at x = a.
1. Use the limit applet(RS4) to find the following given
1 3  x  1

f ( x )   2 x 1  x  0
 x2 0  x  2

a) lim f ( x)
x 1
b) f ( 1)
c) Is f(x) continuous at x = –1?
d) lim f ( x )
x 0
b) f (0)
c) Is f(x) continuous at x = 0?
2. Use the limit applet(RS4) to find the following given
2 x  1 1  x  1

f ( x)  4
x 1
3
1 x  3

a) lim f ( x )
x 1
b) f (1)
c) Is f(x) continuous at x = 1?
3. Use the limit applet(RS4) to find the following given
ACOW LIMITS AND CONTINUITY MODULE
 1

f ( x)   ( x  2)
 x

Updated 5/28/2016
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3  x  0
0x2
a) lim f ( x )
x 1
b) f (1)
c) Is f(x) continuous at x = 1?
4. Which of the following is true regarding the function
2 x  1 2  x  0

f ( x)  3
x0
1
0 x2

Use the limit applet(RS4) if it helps.
a) The function is continuous at x = 0.
b) The function is discontinuous at x = 0 because f (0) is not defined.
c) The function is discontinuous at x = 0 because lim f ( x ) does not exist.
x 0
d) The function is discontinuous at x = 0 because lim f ( x)  f (0) .
x 0
5. Use the limit applet(RS4) to find all values of x where f(x) is discontinuous if
 x2
2 x 0
 3
f ( x)   x
0  x 1 .
0.5 x 1  x  4
