Question 1180321: Given the following piecewise function for f(x), solve:
1/2x, x < 1
x^2, x = 1
ln(x), x > 1
a.
lim f(x) =
x→1−
b.lim f(x) =
x→1+
c. f(1) =
d.
lim f(x) =
x→1
e. Is f(x) continuous at x=1?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Let's define three helper functions
g(x) = (1/2)x
h(x) = x^2
j(x) = ln(x)
The piecewise function f(x) will change its identity based on what x is
If x < 1, then f(x) = g(x) = (1/2)x
If x = 1, then f(x) = h(x) = x^2
If x > 1, then f(x) = j(x) = ln(x)
so it all depends on the input
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Part A
As x approaches 1 from the left, we're going to be using g(x) = (1/2)x, since this tied to the x < 1.
Plug in x = 1
g(x) = (1/2)x
g(1) = (1/2)*1
g(1) = 0.5
Therefore, f(x) is approaching 0.5 when x approaches 1 from the left.
Check out the graph below for a visual of what's going on.
Answer: 0.5
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Part B
We're approaching the other side of x = 1 this time. So we'll be using j(x) = ln(x) this time as this function deals with cases when x > 1.
We see that
j(x) = ln(x)
j(1) = ln(1)
j(1) = 0
Meaning f(x) is approaching 0 as x approaches 1 from the right.
Answer: 0
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Part C
When x = 1, we use the middle function h(x) = x^2
h(x) = x^2
h(1) = 1^2
h(1) = 1
Therefore, f(1) = 1 as well.
Answer: 1
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Part D)
The results of parts A and B will be used to determine the answer here.
From part A, we found the left hand limit was 0.5 which contrasts with the right hand limit of 0 (part B)
The left hand limit and right hand limit must be the same in order for the limit to exist. Since they are different values, the limit does not exist.
Answer: Limit does not exist
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Part E
If the limit doesn't exist at a certain x value, then it's certainly not continuous here either. Also, note how the results of parts A,B,C are different as well. They must be the same value to have the function continuous at this x value.
Answer: No, f(x) is not continuous at x = 1
-----------------------------------------------
Graph:
Notes:- There are open holes at the endpoints (1,0.5) and (1,0)
- The limit does not exist at x = 1 because the left and right sides are not approaching the same y value (ie we have a "disconnected" road or bridge). This leads to the function not being continuous here either.
- If the roads connected, but the point was removed from the road, then we'd still have discontinuity. In order to have continuity at x = 1, the red and purple roads must connect and the blue point must also be on those roads.
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