Determining Continuity with the 3-step Checklist
● 3-Step Checklist for continuity at x=c:
1.
f(c) must exist.
lim f ( x ) must exist.
2.
x c
3.
#1 = #2
Example: Let f(x) be the function whose graph is given.
Use the checklist to determine if f(x) is continuous at the given x values.
If not, state which condition failed.
1.
x = –2
(1) f(–2)=–1
(2) lim f ( x )  1
x  2
(3) #1=#2
f(x) is continuous at x = –2.
2.
x = –3
(1) f(–3) DNE
f(x) is not continuous at x = –3, since condition 1 failed.
3.
x=2
(1) f(2)=3
(2) lim f ( x ) DNE
x 2
( lim f ( x )  3 and lim f ( x )  2 , which means lim f ( x)  lim f ( x) ,
x2
x2
x 2
x 2
so lim f ( x ) DNE.)
x 2
f(x) is not continuous at x = 2, since condition 2 failed.
4.
x=4
(1) f(4)=2
(2) lim f ( x )  3
x4
(3) #1
 #2
f(x) is not continuous at x = 4, since condition 3 failed.
Recall: Discontinuities can be either removable (if the discontinuity can be removed by defining or re-defining the point) or
non-removable (if the discontinuity can’t be removed by defining or re-defining the point, such as with “jump” discontinuities
or asymptotes).
5.
Which of the discontinuities above are removable?
x = –3; x = 4
Which of the discontinuities above are non-removable?
x=2
6.
x 1
 x,
.
f ( x)  
1

x
,
x

1

Consider the function
(A)
Use the 3-step checklist to determine continuity at x = 1.
(1)
(2)
f(1) = 1
lim f ( x ) DNE
x1
( lim f ( x )  1  1  0 and lim f ( x )  1 , which means lim f ( x )  lim f ( x ) ,
x 1
x 1
x 1
x 1
so lim f ( x ) DNE.)
x1
f(x) is not continuous at x = 1, since condition 2 failed.
(B)
Is the discontinuity at x = 1 removable or non-removable?
Because the limit from the left is 0 and the limit from the right is 1, the limit does not exist and we have
a “jump” discontinuity, which non-removable.
7.
Determine all discontinuities and decide whether they are removable or non-removable.
2 x, x  0
g ( x)   2
x , x  0

Because g(0) is undefined (failing the first step of the continuity checklist), there is clearly a discontinuity at x=0.

To decide what type, we must look at the lim g ( x) , which because this is a piece-wise function, we must look at
x0
both sides. So,
lim g ( x)  0  0 and lim g ( x)  2  0  0 . So…
2
x 0
x0
Because the left and right-hand limits are both 0, we know the lim g ( x )  0 . Therefore, we now know that we
x 0
have a removable discontinuity at x = 0.
8.
Determine all discontinuities and decide whether they are removable or non-removable.
f ( x) 
x2
x2  4

When we set the denominator = 0, we find that there are two places where f is undefined; therefore, there is
clearly discontinuities at x = +2.

Because f is a rational function and +2 are places where the denominator go to 0, we know there will either be a
vertical asymptote or a removable discontinuity (a little hole) at x = +2. To decide which, we look at if these
values also make the numerator go to 0. So…
Because x = 2 also makes the numerator go to 0, we have a removable discontinuity (a little hole) at x = 2.
Because x = –2 does not make the numerator go to 0, we have a vertical asymptote (which is non-removable)
at x = –2.