First off I want to apologize for bombarding this subforum with my gazillion questions. If my continuous barrage of questions poses a problem just let me know and I'll stop.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

For each value of ε, find a positive value of δ such that the graph of the function leaves the window (a − δ) < x < (a + δ), (b − ε) < y < (b + ε) by the sides and not through the top or bottom.

g(x) = −x^3 + 2

a = 0

b = 2

ε = 0.1, 0.01, 0.001

For ε = 0.1, δ must be less than or equal to what value?

2. Relevant equations

0<abs(x-a)<δ then abs(f(x)-b)<ε

3. The attempt at a solution

abs(−x^3 + 2-2)<ε

abs(−x^3)<.1

1.9<−x^3<2.1

0<abs(x-0)<δ

-δ<x<δ

1.9<−x^3

-(1.9^1/3)>x

−x^3<2.1

x>-(2.1^1/3)

(-2.1^1/3)<=δ<x<δ<=-(1.9^1/3)

δ<=-(1.9^1/3) || δ<=-1.2386

Is this right? I tried to get everything to match of properly, but I'm not sure if I did it correctly. I'm not exactly sure what I even did just now :/ Thanks in advance and sorry for all the questions.

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# Homework Help: Rigorous definition of a limit

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