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Epsilon Delta Proof for this function?
How do I prove this limit?
lim x -> 2 (1/x) = 1/2
1 Answer
- kbLv 75 years ago
Given ε > 0, we need to find δ > 0 such that 0 < |x - 2| < δ ==> |1/x - 1/2| < ε.
To this end, note that
|1/x - 1/2|
= |2 - x|/|2x|
= (1/2) |x - 2| * 1/|x|.
We need to bound x away from 0; so assume |x - 2| < 1 (for instance).
==> -1 < x - 2 < 1
==> 1 < x < 3
==> 1/3 < 1/x < 1.
So, |1/x - 1/2|
= (1/2) |x - 2| * 1/|x|
< (1/2) |x - 2| * 1
= (1/2)|x - 2|.
Here's the proof:
Given ε > 0, let δ = min{1, 2ε}.
Then, 0 < |x - 2| < δ ==> |1/x - 1/2| < (1/2)|x - 2| < (1/2)(2ε) = ε.
I hope this helps!
