Rigorous definition of a limit

[Painguy]

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.

Homework Statement

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?

Homework Equations

0<abs(x-a)<δ then abs(f(x)-b)<ε

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.

 

[mighty2000]

Thank you for your question. I am here to assist and share knowledge, so please do not apologize for asking questions. It is important to have a clear understanding of concepts and problems in order to find the correct solution.

In this case, we are looking for a positive value of δ such that the graph of the function g(x) = −x^3 + 2 does not cross the window (a − δ) < x < (a + δ), (b − ε) < y < (b + ε) by the sides. We are given a = 0, b = 2, and three values of ε: 0.1, 0.01, and 0.001.

To find the value of δ for each ε, we can use the given equation: 0<abs(x-a)<δ then abs(f(x)-b)<ε. Plugging in the values of a and b, we get:

0<abs(x-0)<δ → 0<x<δ

abs(−x^3 + 2-2)<ε → abs(−x^3)<ε

For ε = 0.1, we have:

abs(−x^3)<0.1

Since we want the graph to not cross the window by the sides, we can set up the following inequalities:

0<x<δ and abs(−x^3)<0.1

Solving for x, we get:

x<δ and −0.1<x^3<0.1

Taking the cube root of both sides, we get:

−0.4641<x<0.4641

Therefore, for ε = 0.1, δ must be less than or equal to 0.4641.

Similarly, for ε = 0.01, we get:

−0.0963<x<0.0963

Therefore, for ε = 0.01, δ must be less than or equal to 0.0963.

For ε = 0.001, we get:

−0.0099<x<0.0099

Therefore, for ε = 0.001, δ must be less than or equal to 0.0099.

I hope this helps clarify the solution. Let me know if you have any further questions. Keep up the good work with your studies!