SOLUTION: 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) =

Algebra.Com
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

-----------------------------------------------
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
-----------------------------------------------
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
-----------------------------------------------
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
-----------------------------------------------
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

-----------------------------------------------
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:

RELATED QUESTIONS

lim f(x)... (answered by ikleyn)
Consider the following function: F(x): 1-x x<-1 x... (answered by stanbon)
A1 {{{f(x)= system( (x^2-1)/(x^2+x-2), if x< 1, 3, if x =1, e^1-x, if... (answered by Aldorozos,MathTherapy)
A1 {{{f(x)= system( ((x^2-1)/(x^2+x-2)), 3, e^1-x, 2x-c (answered by ikleyn)
I have a Calculus problem that asks me to: Find a formula for a function f that... (answered by stanbon)
Consider the following split domain function f(x)= {5x-4, x<1 {2x+2, x>=1 find... (answered by stanbon)
If lim-->1 f(x)/(x-1)(x-2)=-3, then provide a possible function y = f(x) (answered by ikleyn)
lim f(x) x... (answered by Fombitz)
For the following piecewise defined function find f(-1) f(x)={x^2+1,......x<0... (answered by )