Introduction to Mathematical Analysis I - Second Edition

73 Now, 3 x 2 + x 2 x 2 + 1 − 3 2 = | 2 x − 3 | 2 ( 2 x 2 + 1 ) . Therefore, simplifying, 3.5 is equivalent to 1 ε < 2 ( 2 x 2 + 1 ) | 2 x − 3 | . (3.6) We first restrict x to be less than 0, so | 2 x − 3 | > 3. Then, since 4 x 2 3 < 2 ( 2 x 2 + 1 ) | 2 x − 3 | , 3.6 will be guaranteed if 1 / ε < 4 x 2 / 3 or, equivalently p 3 / ( 4 ε ) < | x | . We set c < min { 0 , − p 3 / ( 4 ε ) } . Then, if x < c , we have p 3 / ( 4 ε ) < − x = | x | . Thus, 1 / ε < 2 ( 2 x 2 + 1 ) | 2 x − 3 | and, hence, 3 x 2 + x 2 x 2 + 1 − 3 2 = | 2 x − 3 | 2 ( 2 x 2 + 1 ) < ε . Exercises 3.2.1 Find the following limits: (a) lim x → 2 3 x 2 − 2 x + 5 x − 3 , (b) lim x →− 3 x 2 + 4 x + 3 x 2 − 9 3.2.2 Let f : D → R and let ¯ x is a limit point of D . Prove that if lim x → ¯ x f ( x ) exists, then lim x → ¯ x [ f ( x )] n = [ lim x → ¯ x f ( x )] n , for any n ∈ N . 3.2.3 Find the following limits: (a) lim x → 1 √ x − 1 x 2 − 1 , (b) lim x → 1 x m − 1 x n − 1 , where m , n ∈ N , (c) lim x → 1 n √ x − 1 m √ x − 1 , where m , n ∈ N , m , n ≥ 2, (d) lim x → 1 √ x − 3 √ x x − 1 . 3.2.4 Find the following limits: (a) lim x → ∞ ( 3 √ x 3 + 3 x 2 − √ x 2 + 1 ) . (b) lim x →− ∞ ( 3 √ x 3 + 3 x 2 − √ x 2 + 1 ) . 3.2.5 I Let f : D → R and let ¯ x be a limit point of D . Suppose that | f ( x ) − f ( y ) | ≤ k | x − y | for all x , y ∈ D \ { ¯ x } , where k ≥ 0 is a constant. Prove that lim x → ¯ x f ( x ) exists. 3.2.6 Determine the one-sided limits lim x → 3 + [ x ] and lim x → 3 − [ x ] , where [ x ] denotes the greatest integer that is less than or equal to x .