SOLUTION: Prove lim x^3 = 1 as x approaches 1 using precise definition of limits
This is how far I got and my line of thought not sure if I am making any conceptual errors.
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-> SOLUTION: Prove lim x^3 = 1 as x approaches 1 using precise definition of limits
This is how far I got and my line of thought not sure if I am making any conceptual errors.
Basically t
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Question 1163650: Prove lim x^3 = 1 as x approaches 1 using precise definition of limits
This is how far I got and my line of thought not sure if I am making any conceptual errors.
Basically the question wants me to show that whenever they give me a e>0 value, I can return a &>0 value whereby
0<|x-1|<& --> |x^3 - 1|
I tried to express the right hand inequality in terms of |x-1| so that I can relate to the left hand inequality.
|(x-1)(x^2 + x + 1)|
|(x-1)((x-1)^2 + 3(x-1) - 3)|
If I were to modulus all the x-1, I will get this inequality
||x-1|(|x-1|^2 + 3|x-1| - 3)| >= |(x-1)((x-1)^2 + 3(x-1) - 3)|
And if I were to convert it further to & terms,
&(&^2 + 3& - 3) > ||x-1|(|x-1|^2 + 3|x-1| - 3)| < e
&(&^2 + 3& -3) < e
So from here I am not too sure how to go about it. Please advise Answer by Edwin McCravy(20054) (Show Source):
For any given ε > 0, we must find a δ > 0 such that
|x³ - 1| < ε whenever |x - 1| < δ
|x³ - 1| < ε iff
|(x - 1)(x² + x + 1)| < ε iff
To find the appropriate δ on the interval (0,2), we know that
|x² + x + 1| < 1 (the value of this increasing function when x=0)
So on the interval (0,2),
|(x - 1)(x² + x + 1)| < ε iff
Thus whenever δ < ε, then
|x³ - 1| < ε
thus [PROVED]
Edwin
