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Determine if the next progression have a limit (demostrate it) and determine it
 5 months ago
 5 months ago
Determine if the next progression have a limit (demostrate it) and determine it
 5 months ago
 5 months ago

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osansevieroBest ResponseYou've already chosen the best response.0
\[\left( 1^{n} \right)\left( \frac{ 5n+4 }{ 2n } \right)\]
 5 months ago

tkhunnyBest ResponseYou've already chosen the best response.1
As n increases in the positive direction, \(\dfrac{5n+4}{2n}\) approaches \(\dfrac{5}{2}\). Follow with your mind as n increases. The little 4 on the end of the numerator becomes less and less significant. The terms do NOT approach zero.
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.0
oh...I think I see it now...but how can i demostrate if it is it's limit?
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.0
but there is also the 1^n
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.0
So it approaches 5/2 and 5/2 ?
 5 months ago

tkhunnyBest ResponseYou've already chosen the best response.1
The typical demonstration is a division by n. For n > 0, \(\dfrac{5n+4}{2n} = \dfrac{5 + \dfrac{4}{n}}{2}\). In this form, it is relatively obvious that the limit it 5/2 as n increases. The FIRST criterion for convergence is terms that approach ZERO. Nothing else will do. These terms do not approach zero, therefore, we do not care about the alternating sign. If the terms approach zero, THEN we'll worry about the sign.
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.0
what I mean is that there is a (1^n) multiplying all of that...so 5/2 is also a limit
 5 months ago

tkhunnyBest ResponseYou've already chosen the best response.1
No, this is not a limit. Limits come alone, not in pairs. The terms, without the sign, approach 5/2. I may have stated that carelessly, before. The actual terms, including the sign, do not have a limit. They is oscillating.
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.0
So for this there isnt a limit, neither \[\frac{ 1 }{ 2 },2^{2}, \frac{ 1 }{ 2^{3} }\]
 5 months ago

tkhunnyBest ResponseYou've already chosen the best response.1
Seriously? An alternating sign in the exponent? No. No limit.
 5 months ago

osansevieroBest ResponseYou've already chosen the best response.0
I thought so, thanks :)
 5 months ago
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