Finding Limits Using Formal Definition

[TommG]

Homework Statement

Use formal definition of limits
Find L = lim x→ c f(x). Then find a number δ > 0 such for all x

f(x) = 3 - 2x
c = 3
ε = 0.02

The Attempt at a Solution

lim_x→3 3 -2x

lim_x→3 3 - lim_x→3 2x

3 - 2(3) = -3
L = -3
I am not sure how to find delta

 

[LCKurtz]

Homework Statement

Use formal definition of limits
Find L = lim x→ c f(x). Then find a number δ > 0 such for all x

f(x) = 3 - 2x
c = 3
ε = 0.02

The Attempt at a Solution

lim_x→3 3 -2x

lim_x→3 3 - lim_x→3 2x

3 - 2(3) = -3
L = -3



I am not sure how to find delta

Ask yourself how close $x$ needs to be to $3$ so that $|f(x)-L|<\epsilon$ or, for your problem, $|(3-2x) - (-3)|<.02$.

 

[lurflurf]

|f(x)-L|<ε
when
|3-x|<δ
write
|f(x)-L|
in terms of
|3-x|

 

[TommG]

Ask yourself how close $x$ needs to be to $3$ so that $|f(x)-L|<\epsilon$ or, for your problem, $|(3-2x) - (-3)|<.02$.

ok so i take

-0.02 < (3-2x)-(-3) < 0.02
-0.02 < 6-2x < 0.02
-6.02 < -2x < -5.98
3.01 > x > -2.99
(-2.99,3.01)

-2.99 - 3 = -5.99
3.01 - 3 = 0.1

so since δ > 0
δ = 0.1

matches answer in book

thank you for your help

 

[HallsofIvy]

ok so i take

-0.02 < (3-2x)-(-3) < 0.02
-0.02 < 6-2x < 0.02

Notice that, at this point, you could say
-0.02 < 2(3- x)< 0.02
-0.01< 3- x< 0.01 so that |x- 3|< 0.01

-6.02 < -2x < -5.98
3.01 > x > -2.99
(-2.99,3.01)

-2.99 - 3 = -5.99
3.01 - 3 = 0.1

so since δ > 0
δ = 0.1

matches answer in book

thank you for your help

 

[LCKurtz]

ok so i take

-0.02 < (3-2x)-(-3) < 0.02
-0.02 < 6-2x < 0.02
-6.02 < -2x < -5.98
3.01 > x > -2.99
(-2.99,3.01)

-2.99 - 3 = -5.99
3.01 - 3 = 0.1

so since δ > 0
δ = 0.1

matches answer in book

thank you for your help

That's good. But you could write it much neater:$$
|6-2x| <.02$$ $$
2|3-x| <.02$$ $$
|3-x| <.01$$The steps are reversible so $\delta=.01$.

 

[LCKurtz]

-2.99 - 3 = -5.99
3.01 - 3 = 0.1

Also note that should be -2.99 - (-3) = .01. And .1 should be .01.