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 x such that 0 < |x c| < p h(x) < f(x) < g(x). If limxc h(x) = H and limxc g(x) = G and limxc f(x) = F and H < G then H < F < G. 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 > 0. Since limxc g(x) = G it follows from Definition 2.2.1 that there is some 1 such that |g(x)G| < for all x such that 0 < |xc| < 1. Similarly, since limxc f(x) = F there is some 2 such that |f(x) F| < for all x such that 0 < |x c| < 2. But then let x be such that 0 < |x c| < min{p, 1, 2} and observe that f(x) > F = F (F G)/2 = F/2 + G/2 = G + (F G)/2 = G + > g(x) and this contradicts that f(x) < g(x) since 0 < |x c| < 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 limxc h(x) = H there is some 1 such that |h(x) H| < for all x such that 0 < |x c| < 1 and, since limxc f(x) = F there is some 2 such that |f(x) F| < for all x such that 0 < |x c| < 2. But then, once again, let x be such that 0 < |x c| < min{p, 1, 2} and observe that f(x) < F = F + (H F)/2 = F/2 + H/2 = H (H F)/2 = H < h(x) and this contradicts that h(x) < f(x) since 0 < |x c| < p. The contradiction in both cases establishes the assertion.

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.