Prove limx->1 (3-x)=2 using Delta Epsilon.?
Update:
Can someone verify if my proof is actually a proof/correct? (let E= epsilon and D=delta)
What I did:
If E>0 and D=E then
0<abs(x-c)<D=E
abs(x-c)<E
abs(x-1)<E
abs(1-x)<E
abs(3-x-2)<E
abs(f(x)-L)<E
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Answers & Comments
Verified answer
Given ε > 0, we need to find δ > 0 such that
0 < |x - 1| < δ ==> |(3 - x) - 2| < ε.
Note that
|(3 - x) - 2| = |1 - x| = |-(x - 1)| = |x - 1|.
So given ε > 0, let δ = ε.
Then, 0 < |x - 1| < δ ==> |(3 - x) - 2| = |x - 1| < δ = ε.
I hope this helps!