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Is my epsilon delta proof format correct?
Prove lim x^2 as x -> 1 = 1
After doing scratch work, is the format of this proof correct?
Proof: GIven ε > 0, Let δ = min(1 , ε / 5).
If |x - 1| < δ
3|x - 1| < 3δ
|x+1||x-1| < 3|x-1| < 3δ,
then |x^2 - 1| < 3(ε / 3)
|x^2 - 1| < ε
Q.E.D.
1 Answer
- MorewoodLv 75 years ago
Looks good, aside from a small typo: δ = min(1 , ε/3).
Most profs would be happy with that.
Some students would appreciate a small clarification inserted in the middle.
Assuming that |x - 1| < δ = min(1, ε/3)
The triangle inequality gives:
|x+1| = | (x-1) + 2 | ≤ |x-1| + |2| < δ + 2 ≤ 3
(which telescopes to |x+1| < 3 )
so that
|x² - 1| = |x+1||x-1| < 3|x-1| < 3δ ≤ 3(ε/3) = ε
(which telescopes to |x²-1| < ε )
proving the desired limit.
