Introduction to Mathematical Analysis I - Second Edition

152 Solutions and Hints for Selected Exercises Similar problems: 1. Determine all a ∈ R at which lim x → a f ( x ) exists, where f ( x ) = ( x 2 , if x ∈ Q ; x + 2 , if x ∈ x 6∈ Q . 2. Consider the function f ( x ) = ( x 2 + 1 , if x ∈ Q ; − x , if x 6∈ Q . Prove that f does not have a limit at any a ∈ R . SECTION 3.2 Exercise 3.2.5 . The given condition implies that if both x 1 and x 2 are close to ¯ x , then they are close to each other and, hence, f ( x 1 ) and f ( x 2 ) are close to each other. This suggests the use of the Cauchy criterion for limit to solve the problem. Given any ε > 0, choose δ = ε 2 ( k + 1 ) . If x 1 , x 2 ∈ D \ { ¯ x } with | x 1 − ¯ x | < δ and | x 2 − ¯ x | < δ , then | f ( x 1 ) − f ( x 2 ) | ≤ k | x 1 − x 2 | ≤ k ( | x 1 − ¯ x | + | x 2 − ¯ x | ) < k ( δ + δ ) = 2 k ε 2 ( k + 1 ) < ε . Therefore, lim x → ¯ x f ( x ) exists. SECTION 3.3 Exercise 3.3.8 . (a) Observe that f ( a ) = g ( a ) = h ( a ) and, hence, | f ( x ) − f ( a ) | = ( | g ( x ) − g ( a ) | , if x ∈ Q ∩ [ 0 , 1 ] ; | h ( x ) − h ( a ) | , if x ∈ Q c ∩ [ 0 , 1 ] . It follows that | f ( x ) − f ( a ) | ≤ | g ( x ) − g ( a ) | + | h ( x ) − h ( a ) | for all x ∈ [ 0 , 1 ] . Therefore, lim x → a f ( x ) = f ( a ) and, so, f is continuous at a . (b) Apply part (a). Exercise 3.3.9 . At any irrational number a ∈ ( 0 , 1 ] , we have f ( a ) = 0. If x is near a and x is irrational, it is obvious that f ( x ) = 0 is near f ( a ) . In the case when x is near a and x is rational, f ( x ) = 1 / q where p , q ∈ N . We will see in part (a) that for any ε > 0, there is only a finite number of x ∈ ( 0 , 1 ] such that f ( x ) ≥ ε . So f ( x ) is close to f ( a ) for all x ∈ ( 0 , 1 ] except for a finite number of x ∈ Q . Since a is irrational, we can choose a sufficiently small neighborhood of a to void such x . (a) For any ε > 0, A ε = { x ∈ ( 0 , 1 ] : f ( x ) ≥ ε } = x = p q ∈ Q : f ( x ) = 1 q ≥ ε = x = p q ∈ Q : q ≤ 1 ε .