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Math 301 Problem Set 10 Solutions

# Math 301 Problem Set 10 Solutions - Math 301/ENAS 513...

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Math 301/ENAS 513: Homework 10 Solution Triet M. Le Page 104: 2: f is real on R and | f ( x ) | < M for all x R . Then for any x = y , the mean value theorem implies f ( x ) - f ( y ) x - y = f ( c ) , for some c between x and y. Hence, for any > 0, | f ( x ) - f ( y ) | = | f ( c ) || x - y | ≤ M | x - y | < , whenever, | x - y | < δ = /M . Thus, f is uniformly continuous on R . Page 104: 3a: Fix > 0. Since f ( x ) 0 as | x | → ∞ , there exists N > 0 such that | f ( x ) | < / 2 , if x / [ - N, N ] . f is obviously continuous on [ - ( N + 1) , N + 1] which is compact, hence it is uniformly continuous. This implies there exists 0 < δ < 1 such that | f ( x ) - f ( y ) | < , whenever x, y [ - ( N + 1) , N + 1] and | x - y | < δ. Now, for any x, y R such that | x - y | < δ . if both x, y [ - ( N + 1) , N + 1], then by the above equation, | f ( x ) - f ( y ) | < . On the otherhand if either x / [ - ( N +1) , N +1] and/or y / [ - ( N +1) , N +1], the condition | x - y | < δ < 1 implies that both x, y / [ - N, N ]. This implies, | f ( x ) - f ( y ) | ≤ | f ( x ) | + | f ( y ) | < . Thus, in all cases, | f ( x ) - f ( y ) | < whenever | x - y | < δ , which shows that f is uniformly continuous on R . Page 104: 3b: Consider f ( x ) = sin( x 4 ) / ( x 2 + 1). Clearly, f is continuous and differen- tiable on R

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