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Problem Set #5 - Chapter 5

January 29 - February 4, 2011

Spivak 3rd Edition: 3, 12, 13, 18, 26 Exercise 1. In each of the following cases, find δ such that |f (x) − l| <  for all x satisfying 0 < |x − a| < δ. (i) f (x) = x4 ; l = a4 . (ii) f (x) =

1 ; a = 1, l = 1. x

1 ; a = 1, l = 2. x x (iv) f (x) = ; a = 0, l = 0. 1 + sin2 x p (v) f (x) = |x|; a = 0, l = 0. √ (vi) f (x) = x; a = 1, l = 1.

(iii) f (x) = x4 +

Exercise 2. (i) Suppose that f (x) ≤ g(x) for all x. Prove that lim f (x) ≤ lim g(x), provided that these x→a x→a limits exist. (ii) How can the hypotheses be weakened? (iii) If f (x) < g(x) for all x, does it necessarily follow that lim f (x) < lim g(x)? x→a

x→a

Exercise 3. Suppose that f (x) ≤ g(x) ≤ h(x) and that lim f (x) = lim h(x). Prove that lim g(x) exists, x→a

x→a

x→a

and that lim g(x) = lim f (x) = lim h(x). (Draw a picture!). x→a

x→a

x→a

Exercise 4. Prove that if lim f (x) = l, then there is a number δ > 0 and a number M such that |f (x)| < M x→a

if 0 < |x−a| < δ.(What does this mean pictorially?) Hint: Why does it suffice to prove that l−1 < f (x) < l+1 for 0 < |x − a| < δ? Exercise 5. Give examples to show that the following definitions of lim f (x) = l are not correct. x→a

(i) For all δ > 0 there is an  > 0 such that if 0 < |x − a| < δ, then |f (x) − l| < . (ii) For all  > 0 there is a δ > 0 such that if |f (x) − l| < , then 0 < |x − a| < δ.

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