Finding Delta for a Given Epsilon and Limit: 3-2x, x0=3, E=.02

[Not An Einstein]

Given a function f(x), a point x0, and a positive number E (epsilon), write the limit then find delta>0 such that for all x 0< |x-x0| < delta -> |f(x)-L| < E
f(x) = 3-2x, x0=3, E=.02
Here is my attempt:
Lim (3-2x) as x->3 = -3
-.02 < |3-2x - 3| <.02
-.02 < |-2x| < .02
.01 > x > -.01
-2.99 > x-3 > -3.01
-2.99 > |x-3|
Delta= -2.99
Is this right? I'm really confused, any help would be greatly appreciated, even just an explanation. Thanks.

 

[jedishrfu]

There's a Khan Academy video on the topic:

https://www.khanacademy.org/math/di...roving-a-limit-using-epsilon-delta-definition

 

[Dick]

Given a function f(x), a point x0, and a positive number E (epsilon), write the limit then find delta>0 such that for all x 0< |x-x0| < delta -> |f(x)-L| < E
f(x) = 3-2x, x0=3, E=.02
Here is my attempt:
Lim (3-2x) as x->3 = -3
-.02 < |3-2x - 3| <.02
-.02 < |-2x| < .02
.01 > x > -.01
-2.99 > x-3 > -3.01
-2.99 > |x-3|
Delta= -2.99
Is this right? I'm really confused, any help would be greatly appreciated, even just an explanation. Thanks.

You started going off track when you wrote |f(x)-L|<E. Since L is -3 that becomes |(3-2x)-(-3)|=|3-2x+3|=|6-2x|<0.02. Try it again from there.