# homework_4 - Homework 4 8.4. Suppose that there exists an M...

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8.4. Suppose that there exists an M > 0 such that | t n | ≤ M for all n N . To prove that lim( s n t n ) = 0 let ± > 0 and then feed ± M into the deﬁnition of lim s n = 0 to obtain N such that n > N implies | s n | < ± M . Then, n > N implies | t n s n | = | t n || s n | < M ± M = ±. 8.10. Let s = lim n →∞ s n . Then, s - a > 0. Taking ± = 1 2 ( s - a ), we have that there exists an N such that n > N implies | s n - s | < 1 2 ( s - a ) . Since this implies - 1 2 ( s - a ) < ( s n - s ) < 1 2 ( s - a ) we have s n > s - 1 2 ( s - a ) . Then, s n > s - 1 2 ( s - a ) = 1 2 ( s + a ) = a + 1 2 ( s - a ) > a since s - a > 0. 9.2. (a) Since the sequences x n and y n converge, we have by the addition theorem for limits that lim( x n + y n ) = lim( x n ) + lim( y n ) = 3 + 7 = 10 . (b) Since the two sequences converge, and all of the y n are non-zero and lim y n 6 = 0, we have by the division theorem for limits that lim( 3 y n - x n y 2 n ) = lim(3 y n - x n ) lim( y 2 n ) and now we can use the multiplication theorem for limits to reduce the

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homework_4 - Homework 4 8.4. Suppose that there exists an M...

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