If | x + 3 | ≥ 10 , then 1)xϵ(−13,7] 2)xϵ(−13,7) 3)xϵ(−∞,−13)∪(7,∞) 4)x ϵ(−∞,−13)∪[7,∞)
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Let f(x) = 1 if x is rational and f(x) = 0 if x is irrational. ... Then there exists δ > 0 such that for all y ∈ R with |y − x| < δ, |f(y) − f(x)| < 1. This means that for any y ∈ (x − δ, x + δ), |f(y) − f(x)| < 1. If x is rational, there exists y ∈ (x − δ, x + δ) which is irrational, and |f(y) − f(x)| = |0 − 1|
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