Calculus- Early Transcendentals, 2021a

106 Limits
Example 3.44: Continuous at a Point
What value of c will make the following function f (x) continuous at 2?
⎧
x ⎪⎨
2 − x − 2
if x ≠ 2
f (x)= x − 2
⎪⎩
c if x = 2
Solution. In order to be continuous at 2 we require
lim f (x)= f (2)
x→2
to hold. We use the three part definition listed previously to check this.
1. First, f (2)=c, andc is some real number. Thus, f (2) is defined.
2. Now, we must evaluate the limit. Rather than computing both one-sided limits, we just compute the
limit directly. For x close to 2 (but not equal to 2) we can replace f (x) with x2 −x−2
x−2
to get:
lim f (x)=lim x 2 − x − 2 (x − 2)(x + 1)
= lim
= lim(x + 1)=3.
x→2 x→2 x − 2 x→2 x − 2 x→2
Therefore the limit exists and equals 3.
3. Finally, for f to be continuous at 2, we need that the numbers in the first two items to be equal.
Therefore, we require c = 3. Thus, when c = 3, f (x) is continuous at 2, for any other value of c, f (x) is
discontinuous at 2.
♣
For continuity on a closed interval, we consider the one-sided limits of a function. Recall that x → a −
means x approaches a from values less than a.