Theorems on limits of sequences ove that lim ab AB. 2.8. Prove that if lim u, exists, it must be unique.As.b- -A We must show that if lim u, = l, and lim u, =,, then l =2. n00 n00 By hypothesis, given any e > 0 we can find N such that l4, – 4| N, u, - 4| N Then 3= 3 + 3>|1- "n| +|"n– /|5|7 – "n + "n – '/%=|7- | Haac i.e., |4, -4| is less than any positive e (however small) and so must be zero. Thus, l

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I have to explain each step of the solved problems in the picture.
Theorems on limits of sequences prove that lim ab=AB.
Prove that if lim u, exists, it must be unique. As -A
2.8.
We must show that if lim u, = 1, and lim u,= 2, then l = 2.
By hypothesis, given any e > 0 we can find N such that
4.-4|<e when n> N, u, -4|<e
when n > N
Then
14-4|=14-4, +u, -4|<4-u,+|4, – 4|<;e +e =ɛ
%3D
Iarger c
i.e., |1, -4| is less than any positive e (however small) and so must be zero. Thus, 4 = l2.
Transcribed Image Text:Theorems on limits of sequences prove that lim ab=AB. Prove that if lim u, exists, it must be unique. As -A 2.8. We must show that if lim u, = 1, and lim u,= 2, then l = 2. By hypothesis, given any e > 0 we can find N such that 4.-4|<e when n> N, u, -4|<e when n > N Then 14-4|=14-4, +u, -4|<4-u,+|4, – 4|<;e +e =ɛ %3D Iarger c i.e., |1, -4| is less than any positive e (however small) and so must be zero. Thus, 4 = l2.
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