Let {Sn} be a sequence such that Sn > 0 for all n and Sn --> s , s > 0.(adsbygoogle = window.adsbygoogle || []).push({});

Prove that log (Sn) --> log (s) using the definition of convergence.

Also, we can use the following fact:

log(1 + x) <= x for all x > -1

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My attempt:

Point 1:

note that log(Sn) <= Sn - 1 and log(s) <= s - 1

which implies that

log(Sn) - log(s) <= (Sn-1) - (s -1),

Point 2:

Sn --> s means

for all epsilon, there exist N such that for all n > N implies

|Sn - s| < epsilon.

Thus, it follows from Point 1 and Point 2 that,

|log Sn| - |log s| <= |Sn-1| - |s-1| <= |(Sn-1) - (s - 1)| <= |Sn - s| < epsilon

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but |log Sn| - |log s| is not |log Sn - log s|.

so I am sort of stuck at this point.

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# Solving a limit problem using the definition of convergence

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