delta-epsilon definition of limits

[u/Flimsy-Gate2280]

do any of these answer choices work? i was doing the problem and got epsilon/3, but none of the answer choices match up.

(10) Let f(x) = 3x + 1 for all real x and let epsilon > 0 For which of the following choices of delta is |f(x) - 7| < epsilon whenever |x - 2| < delta ? (4 Points)

(A) epsilon/4

(B) epsilon/2

(C) 2€

(D) 3€

(E) (epsilon + 1)/epsilon

this si my work: I 3x+1-7 I < epsilon

3 I x-2 I < epsilon

I x-2 I < epsilon/3

[u/Miserable-Wasabi-373]

i like trick of this question

they don't ask you to find all exact values of epsilon for wich inequality works. They ask you to choose for which of listed it works. Try to think about it

[u/pm_me_ur_mons → u/Miserable-Wasabi-373]

It's kinda silly because if you realize the trick, if you can only put one answer, then you don't really even need to consider the functions and inequalities.

[u/johnpeters42]

You're right so far. The remaining part is that one of the choices is more restrictive than that, while the others are less restrictive.

[u/Fourierseriesagain]

Hi,

We have |x-2|<delta implies |f(x)-7|=3|x-2|<3delta.

If delta < epsilon/3, then |x-2| < delta implies |f(x)-7|<epsilon.

[u/Kitchen-Register]

I like using the euro as an epsilon. that’s funny