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1 vote
Piece function:

x^2-5: x>3
-2x+8: x less than 3
5: x=3

a)lim 3+ -----> I got 1
b) lim 3- -----> I got 2
c) lim 3 -----> I got 5
d) f(3) -----> I put 5
e) Therefore is f(x) continuous at x = 3 explain.
my answer: No because the limits from both sides are different

User Dawied
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1 Answer

1 vote

First, make the table so you can sketch the graph:

x | y

5 | 20

4 | 9 x² - 5 ; x > 3

3 | 4 (open dot) from the right

3 | 5 5 ; x = 3 ⇒ at x = 3

3 | 2 (open dot) ⇒ from the left

2 | 4 -2x + 8 ; x < 3

1 | 6


Next, look at the graph (or table) to find the limits:

lim 3⁺ = 4 as x approaches 3 from the right, y approaches 4

lim 3⁻ = 2 as x approaches 3 from the left, y approaches 2

lim 3 = DNE lim 3⁺ ≠ lim 3⁻ so the limit does not exist

f(3) = 5 when x = 3, y = 5

f(x) is NOT continuous at x = 3 because lim 3 ≠ f(3)

User Jacke
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8.8k points