121616949-math.250

# 121616949-math.250 - f x = sin xπ-1 1 1 2 3 4 5 6 7 8 •...

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236 Chapter 10 Sequences and Series and closer to a single value, but take on all values between - 1 and 1 over and over. In general, whenever you want to know lim n →∞ f ( n ) you should ﬁrst attempt to compute lim x →∞ f ( x ), since if the latter exists it is also equal to the ﬁrst limit. But if for some reason lim x →∞ f ( x ) does not exist, it may still be true that lim n →∞ f ( n ) exists, but that you’ll have to ﬁgure out another way to compute it. It is occasionally useful to think of the graph of a sequence. Since the function is deﬁned only for integer values, the graph is just a sequence of dots. In ﬁgure 10.1 we see the graphs of two sequences and the graphs of the corresponding real functions. 0 1 2 3 4 5 0 5 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .......................................................................................... f ( x ) = 1 /x 0 1 2 3 4 5 0 5 10 f ( n ) = 1 /n - 1 0 1 . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...
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Unformatted text preview: . . . . . . . . . . . . . . . . . . . . f ( x ) = sin( xπ )-1 1 1 2 3 4 5 6 7 8 • • • • • • • • • f ( n ) = sin( nπ ) Figure 10.1 Graphs of sequences and their corresponding real functions. Not surprisingly, the properties of limits of real functions translate into properties of sequences quite easily. Theorem 2.7 about limits becomes THEOREM 10.2 Suppose that lim n →∞ a n = L and lim n →∞ b n = M and k is some constant. Then lim n →∞ ka n = k lim n →∞ a n = kL lim n →∞ ( a n + b n ) = lim n →∞ a n + lim n →∞ b n = L + M lim n →∞ ( a n-b n ) = lim n →∞ a n-lim n →∞ b n = L-M lim n →∞ ( a n b n ) = lim n →∞ a n · lim n →∞ b n = LM lim n →∞ a n b n = lim n →∞ a n lim x → a b n = L M , if M is not 0...
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