Let f: R → R be the function f(x) = x². Fill in the blanks in the following proof. Claim: f is continuous at 1 E R. Proof: Fix € We have |x² - 1| = |x + 1||x-1|. If |x-1|< 1 then by the triangle inequality Set 6 = ≤|x-1| +2< {1, €/3}. Then for |x-1| < 6 we have |x + 1| < |x-1| < From this we find x² - 1] = |x+2| 1 3 2 min x+1| max 6 ≤ 3(e/3) = e. 0

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