Introduction to Mathematical Analysis I - Second Edition

123 (c) lim x → ∞ π 2 − arctan x ln ( 1 + 1 x ) . (d) lim x → ∞ √ xe − x . ( Hint: first rewrite as a quotient.) 4.4.4 Prove that the following functions are differentiable at 1 and -1. (a) f ( x ) =   x 2 e − x 2 , if | x | ≤ 1; 1 e , if | x | > 1 . (b) f ( x ) =   arctan x , if | x | ≤ 1; π 4 sign x + x − 1 2 , if | x | > 1 . 4.4.5 B Let P ( x ) be a polynomial. Prove that lim x → ∞ P ( x ) e − x = 0 . 4.4.6 I Consider the function f ( x ) = ( e − 1 x 2 , if x 6 = 0; 0 , if x = 0 . Prove that f ∈ C n ( R ) for every n ∈ N . 4.5 TAYLOR’S THEOREM In this section, we prove a result that lets us approximate differentiable functions by polynomials. Theorem 4.5.1 — Taylor’s Theorem. Let n be a positive integer. Suppose f : [ a , b ] → R is a function such that f ( n ) is continuous on [ a , b ] , and f ( n + 1 ) ( x ) exists for all x ∈ ( a , b ) . Let ¯ x ∈ [ a , b ] . Then for any x ∈ [ a , b ] with x 6 = ¯ x , there exists a number c in between ¯ x and x such that f ( x ) = P n ( x )+ f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − ¯ x ) n + 1 , where P n ( x ) = n ∑ k = 0 f ( k ) ( ¯ x ) k ! ( x − ¯ x ) k . Proof: Let ¯ x be as in the statement and let us fix x 6 = ¯ x . Since x − ¯ x 6 = 0, there exists a number λ ∈ R such that f ( x ) = P n ( x )+ λ ( n + 1 ) ! ( x − ¯ x ) n + 1 . We will now show that λ = f ( n + 1 ) ( c ) ,