Verified answer

I'm sorry this is not really about limit. It is about a split function (or piecewise-defined functions).

Given

lim(x→1) f(x) = 3

δ > 0

|f(x)-3| < ε

0 < |x-1| < δ

From the graph f(x) is

f(x) = { 2x + 1 for 0 < x ≤ 1

........ { (1/4)x + 11/4 for 1 ≤ x

When x = 1

f(x) = 3 thus

f(x) - 3 = 0

(a) When ε = 2

|f(x) - 3| < 2

This is equivalent to

-2 < f(x) - 3 < 2

3 - 2 < f(x) < 2 + 3

1 < f(x) < 5

f(x) is maximum when f(x) < 5

(1/4)x + 11/4 < 5

(1/4)x < 5 - 11/4 = 9/4

x < 9*4/4 = 9

9 is greater than 1 therefore it satisfies the second equation

|x - 1| < 9 - 1 = 8

δ = 8

(b) When ε = 1

|f(x) - 3| < 1

This is equivalent to

-1 < f(x) - 3 < 1

3 - 1 < f(x) < 1 + 3

2 < f(x) < 4

f(x) is maximum when f(x) < 4

(1/4)x + 11/4 < 4

(1/4)x < 4 - 11/4 = 5/4

x < 5*4/4 = 5

5 is greater than 1 therefore it satisfies the second equation

|x - 1| < 5 - 1 = 4

δ = 4

(c)

When x=1

f(x) - 3 = 0

Thus when ε ≤ 0 the first equation applies and when ε ≥ 0 the second equation.

For ε ≤ 0

|2x + 1 - 3| < ε ... absolute value of f(x)-1 is less than ε

-ε < 2x - 2 < ε ... that's equivalent to this

2-ε < 2x < 2+ε

1-ε/2 < x < 1+ε/2

|x - 1| < 1+ε/2 - 1 = ε/2 ... the maximum value of |x-1|

δ = ε/2

For ε≥0

|(1/4)x + 11/4 - 1| < ε ... absolute value of f(x)-1 is less than ε

-ε < (1/4)x + 7/4 < ε ... that's equivalent to this

-4ε < x + 7 < 4ε

-4ε-7 < x < 4ε-7

|x - 1| < 4ε-7 - 1 = 4ε - 8 ... the maximum value of |x-1|

δ = 4ε - 8

I'm sorry I did not get the answers provided.