Question 2 What is wrong with the argument claiming to prove the following assertion. You should...

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Question 2 What is wrong with the argument claiming to prove the following assertion. You should not only find the error, but explain why it is an error providing an example. Assertion: For any two numbers a and 3 define max(a. 3) to be the largest of a er 3. If f and g are continuous at: then so is the function max(f.9) defined by max(f.g)(x) = max(f(r).g(z)). Proof: According to Definition 2.4.1 it must be shown that lim max(f.g)(x) = max(f.g)(=). According to Definition 2.2.1, in order to show this, it must be shown that for every > 0 there is some 6>0 such that max(f.g)(x)-max(f.g)(=) < whenever 0 <=-2 <6. So let e> 0 be arbitrary. By the definition of max(f.g)(=)= max(f(=).g(=)) it must be the case that either max(f.g)(=) = f(=) or max(f.g)(=) = g(=) Only the first case will be considered since the argument in the second case is identical. If max (f. 9) (=) = f(=) then there must be some non-empty interval (=-p.=+p) such that max (f.g)(x) = f(x) (1) Juris Steprins 2022. Permission to make digital or hand copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. All other use is prohibited. whenever 0 <-<6". Now let & be the minimum of 6 and p. Then if 0<-< it follows that (2) for all z such that :-p<r<:+p. Since f is continuous at : it follows by Definition 2.4.1 that lim f(x) = f(=) and by Definition 2.2.1 that there is some >0 such that If(x)-f(=)| << as required. =-p<z<=+p and so from (1) that max(f.g)(r) = f(r). But also 0<-<6 and so it follows from (3) that f(x)-f(=) < Combining (1), (2) and (3) yields that [max(f.g)(r)-max(f.9)(=) < Question 3 Find the two errors in the following argument purporting to prove the following version of Theorem 2.5.1. As always, an explanation is required. (One of the errors is an error of logic, while the other error is more of a typo.) Assertion: Let p>0. Suppose that for all z such that 0</z-c<p h(x) < f(x) < g(x). If lim h(r) = H and lim g(x)=G and lim f(x)= F and H<G then H< F<G. E-K THE Proof: Proceed to prove this by contradiction. So suppose that it fails to be the case that H< F<G. There are two possibilities. The first is that G<F. In this case let <= (F-G)/2 and note that e > 0. Since lim g(x)= G it follows from Definition 2.2.1 that there is some 5, such that (g(x)-G]<for all z such that 0<r-cl<8₁. Similarly, since lim f(r) - F there is some 8, such that f(x)-F<for all z such that 0<\-c|< 8₂. But then let z be such that 0<|r-c<min{p, 6₂,6₂) and observe that f(x) > F-<=F-(F-G)/2=F/2+G/2=G+(F-G)/2=G+*>g(z) and this contradicts that f(x) < g(x) since 0<-<p The other possibility is that F<H. The argument in this case is similar to that of the first. Let <= (H-F)/2 and note that >0 as before. Again using Definition 2.2.1 and lim A(z)= H there is some 6, such that h(r)-H|< for all z such that 0<jr-c< 6₂ and, since lim f(2)= F there is some & such that f(z)-F< for all z such that 0<-<6₂. But then, once again, let z be such that 0<r- <min(p.6₂.6₂} and observe that f(x) <F=e=F+(H-F)/2= F/2+H/2=H-(H-F)/2=H-e<h(z) and this contradicts that h(z) <f(z) since 0</z-d<p The contradiction in both cases establishes the assertion. Question 4 Formulate a correct version of the assertion from Question 3 and prove it. The key here is to correct the error of logic and to realize the small, but critical, change it requires to be made to the statement. Once you have found the typo, that is easily corrected. Question 2 What is wrong with the argument claiming to prove the following assertion. You should not only find the error, but explain why it is an error providing an example. Assertion: For any two numbers a and 3 define max(a. 3) to be the largest of a er 3. If f and g are continuous at: then so is the function max(f.9) defined by max(f.g)(x) = max(f(r).g(z)). Proof: According to Definition 2.4.1 it must be shown that lim max(f.g)(x) = max(f.g)(=). According to Definition 2.2.1, in order to show this, it must be shown that for every > 0 there is some 6>0 such that max(f.g)(x)-max(f.g)(=) < whenever 0 <=-2 <6. So let e> 0 be arbitrary. By the definition of max(f.g)(=)= max(f(=).g(=)) it must be the case that either max(f.g)(=) = f(=) or max(f.g)(=) = g(=) Only the first case will be considered since the argument in the second case is identical. If max (f. 9) (=) = f(=) then there must be some non-empty interval (=-p.=+p) such that max (f.g)(x) = f(x) (1) Juris Steprins 2022. Permission to make digital or hand copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. All other use is prohibited. whenever 0 <-<6". Now let & be the minimum of 6 and p. Then if 0<-< it follows that (2) for all z such that :-p<r<:+p. Since f is continuous at : it follows by Definition 2.4.1 that lim f(x) = f(=) and by Definition 2.2.1 that there is some >0 such that If(x)-f(=)| << as required. =-p<z<=+p and so from (1) that max(f.g)(r) = f(r). But also 0<-<6 and so it follows from (3) that f(x)-f(=) < Combining (1), (2) and (3) yields that [max(f.g)(r)-max(f.9)(=) < Question 3 Find the two errors in the following argument purporting to prove the following version of Theorem 2.5.1. As always, an explanation is required. (One of the errors is an error of logic, while the other error is more of a typo.) Assertion: Let p>0. Suppose that for all z such that 0</z-c<p h(x) < f(x) < g(x). If lim h(r) = H and lim g(x)=G and lim f(x)= F and H<G then H< F<G. E-K THE Proof: Proceed to prove this by contradiction. So suppose that it fails to be the case that H< F<G. There are two possibilities. The first is that G<F. In this case let <= (F-G)/2 and note that e > 0. Since lim g(x)= G it follows from Definition 2.2.1 that there is some 5, such that (g(x)-G]<for all z such that 0<r-cl<8₁. Similarly, since lim f(r) - F there is some 8, such that f(x)-F<for all z such that 0<\-c|< 8₂. But then let z be such that 0<|r-c<min{p, 6₂,6₂) and observe that f(x) > F-<=F-(F-G)/2=F/2+G/2=G+(F-G)/2=G+*>g(z) and this contradicts that f(x) < g(x) since 0<-<p The other possibility is that F<H. The argument in this case is similar to that of the first. Let <= (H-F)/2 and note that >0 as before. Again using Definition 2.2.1 and lim A(z)= H there is some 6, such that h(r)-H|< for all z such that 0<jr-c< 6₂ and, since lim f(2)= F there is some & such that f(z)-F< for all z such that 0<-<6₂. But then, once again, let z be such that 0<r- <min(p.6₂.6₂} and observe that f(x) <F=e=F+(H-F)/2= F/2+H/2=H-(H-F)/2=H-e<h(z) and this contradicts that h(z) <f(z) since 0</z-d<p The contradiction in both cases establishes the assertion. Question 4 Formulate a correct version of the assertion from Question 3 and prove it. The key here is to correct the error of logic and to realize the small, but critical, change it requires to be made to the statement. Once you have found the typo, that is easily corrected.

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