161 SECTION 3.1 Exercise 3.1.11. (a) Observe that when x is near 1/2, f (x) is near 1/2 no matter whether x is rational or irrational. We have ( |x − 1/2|, if x ∈ Q; | f (x) − 1/2| = |1 − x − 1/2|, if x ̸ ∈ Q. Thus, | f (x) − 1/2| = |x − 1/2| for all x ∈ R. Given any ε > 0, choose δ = ε. Then | f (x) − 1/2| < ε whenever |x − 1/2| < δ . Therefore, limx →1/2 f (x)= 1/2. (b) Observe that when x is near 0 and x is rational, f (x) is near 0. However, when f is near 0 and x is irrational, f (x) is near 1. Thus, the given limit does not exists. We justify this using the sequential criterion for limits (Theorem 3.1.2). By contradiction, assume that lim f (x)= ℓ, x→0 where ℓ is a real number. Choose a sequence {rn} of rational numbers that converges to 0, and choose also a sequence {sn} of irrational numbers that converges to 0. Then f (rn)= rn and f (sn)= 1− sn and, hence, ℓ = lim f (rn)= 0 n→∞ and ℓ = lim f (sn)= lim (1 − sn)= 1. n→∞ n→∞ This is a contradiction. (c) By a similar method to part (b), we can show that limx →1 f (x) does not exists. Solving this problem suggests a more general problem as follows. Given two polynomials P and Q, defne the function (P(x), if x ∈ Q; f (x)= Q(x), if x ̸ ∈ Q. If a is a solution of the equation P(x)= Q(x), i.e., P(a)= Q(a), then the limit limx →a f (x) exists and the limit is this common value. For all other points the limit does not exist. Similar problems: 1. Determine all a ∈ R at which limx →a f (x) exists, where ( x2 , if x ∈ Q; f (x)= x + 2, if x ∈ x ̸ ∈ Q. 2. Consider the function ( x2 + 1, if x ∈ Q; f (x)= −x, if x ̸ ∈ Q. Prove that f does not have a limit at any a ∈ R.
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