Math 0420 Homework 3. Scratch Work:For c > 0, according to the definition of limit, we want to find δ to bound the term x c by ɛ.we have: x c.

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1 Math 0420 Homework 3 Eercise 311 Find the it or prove the it does not eit (a) for c 0 Solution: First, we might guess that the it is c Scratch Work:For c > 0, according to the definition of it, we want to find δ to bound the term c by ɛwe have: c = ( + c c) = + c c + c c c = 1 c c < 1 c δ So far we have established that: c < 1 c δ It is sufficient to put the right hand side be ɛ That is 1 c δ = ɛ Solve for δ and get δ = cɛ For c = 0, compute in the same way, we get δ = ɛ 2 Formal Proof: For c > 0, let ɛ > 0, choose δ = cɛ, for any such that 0 < c < δ, c = ( + c c c) = c = 1 c < 1 δ = ɛ + c + c c c c Therefore, by the definition of it,for c > 0, = c For c = 0, take δ = ɛ 2, then for any such that 0 < c < δ, c = 0 < δ 2 = ɛ Therefore, by the definition of it,for c = 0, = c Totally, we have, = c 1

2 (c) 0 2 cos( 1 ) Solution: First, we can guess that the it is 0 Scratch Work: According to the definition of it, we want to find δ for 0 to bound the following term by ɛ, 2 cos( 1 ) 0 = 2 cos( 1 ) = 2 cos( 1 ) 2 = 2 Here we have used the property that cos( 1 ) is bounded by 1 So far we have established that: 2 cos( 1 ) 0 < δ2 We can set the right hand side to be ɛ That is δ 2 = ɛ Solve for δ and give δ = ɛ Formal Proof: Let ɛ > 0, choose δ = ɛ, for any such that 0 < 0 < δ, 2 cos( 1 ) 0 = 2 cos( 1 ) = 2 cos( 1 ) 2 = 2 < δ 2 = ɛ Therefore, by the definition of it, 0 2 cos( 1 ) = 0 (e) 0 sin cos( 1 ) Solution: First, we can guess that the it is 0 Scratch Work: According to the definition of it, we want to find δ for 0 to bound the following term by ɛ, sin cos( 1 ) 0 = sin cos( 1 ) = sin cos( 1 ) sin Here we can applied two inequalities: cos( 1 ) 1 and sin So far we have established that: sin cos( 1 ) 0 < δ 2

3 We want the right hand side to be ɛ That is δ = ɛ Formal Proof: Let ɛ > 0, choose δ = ɛ, for any such that 0 < 0 < δ, sin cos( 1 ) 0 = sin cos( 1 ) = sin cos( 1 ) sin < δ = ɛ Therefore, by the definition of it, Eercise 312 Prove Corollary cos( 1 ) = 0 Corollary 3110 Let S R and c be a cluster point of S Let f : S R be a function And suppose that the it of f() as goes to c eists Suppose that there are two real numbers a and b such that a f() b for all S Then, a f() b Proof: Let { n } be a sequence of numbers from S\{c} such that n c We know, By Corollary 224, a f( n ) b for all n a f( n ) b Since the it of f() eists at c and n c, we know L = f( n ) Therefore, a f() b Eercise 314 Prove Corollary 3112 Corollary 3112 Let S R and c be a cluster point of S Let f : S R and g : S R be functions And suppose that the its of f() and g() as goes to c eist Then (i) (f() + g()) = ( f()) + ( g()) (ii) (f() g()) = ( f()) ( g()) (iii) (f()g()) = ( f())( g()) 3

4 (iv) If g() 0 and g() 0 for all S, then (f() g() ) = f() g() Proof: Assume that f() = L and g() = M for some L and M lemma 317, we can say that for any sequence n such that n c, we have, By f( n) = L and g( n ) = M (i) So for arbitrary sequence n such that n c, we obtain, L + M = f( n ) + g( n ) = (f( n ) + g( n )) The last equation follows proposition 225(i) Now we have established that (f( n ) + g( n )) = L + M for all sequences which converge to c Apply Lemma 317 again (in the other direction), we get, (f() + g()) = L + M (f() + g()) = ( f()) + ( g()) (ii) Similarly as above, for arbitrary sequence n such that n c, we obtain, L M = f( n ) g( n ) = (f( n ) g( n )) The last equality is based on proposition 225(ii) Now we have established that (f( n ) g( n )) = L M for all sequences which converge to c Apply Lemma 317 again (in the other direction), we get, (f() g()) = L M (f() g()) = ( f()) ( g()) (iii) Similarly as above, for arbitrary sequence n such that n c, we obtain, LM = f( n ) g( n ) = (f( n )g( n )) 4

5 The last equality follows proposition 225(iii) Now we have established that (f( n )g( n )) = LM for all sequences which converge to c Apply Lemma 317 again (in the other direction), we get, (f()g()) = LM (f()g()) = ( f())( g()) (iv) Similarly as above, for arbitrary sequence n such that n c, we obtain, g( n ) 0 and M 0, Furthermore, L M = f( n ) g( n ) = f( n ) g( n ) The last equality is based on proposition 225(iv) f( It means n) = L g( n) for all sequences which converge to c M Apply Lemma 317 again (in the other direction), we get, f() g() = L M f() g() = f() g() Eercise 315 Let A S, show that if c is a cluster point of A, then c is a cluster point of S Proof: Let c is a cluster point of A, by the definition of cluster point, we have: Since A S, then ɛ > 0, (c ɛ, c + ɛ) A\{c} So, (c ɛ, c + ɛ) A\{c} (c ɛ, c + ɛ) S\{c} (c ɛ, c + ɛ) S\{c} Therefore, again by the definition of cluster point, c is a cluster point of S 5