161 SECTION 3.1 Exercise 3.1.11. (a) Observe that whenx is near 1/2, f (x) is near 1/2 no matter whether x is rational or irrational. We have | f (x)−1/2| =(|x−1/2|, if x ∈Q; |1−x−1/2|, if x̸ ∈Q. Thus, | f (x)−1/2| =|x−1/2| for all x ∈R. Given anyε >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 whenx is near 0 andx is rational, f (x) is near 0. However, when f is near 0 andx 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 x→0 f (x)=ℓ, whereℓ 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 n→∞ f (rn)=0 and ℓ =lim n→∞ f (sn)=lim n→∞ (1−sn)=1. 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 Pand Q, define the function f (x)=( P(x), if x ∈Q; Q(x), if x̸ ∈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∈Rat which limx →a f (x) exists, where f (x)=( x2, if x ∈Q; x+2, if x ∈x̸ ∈Q. 2. Consider the function f (x)=( x2 +1, if x ∈Q; −x, if x̸ ∈Q. Prove that f does not have a limit at anya∈R.
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