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osanseviero
 one year ago
Determine if the next progression have a limit (demostrate it) and determine it
osanseviero
 one year ago
Determine if the next progression have a limit (demostrate it) and determine it

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

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.1As 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.

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.0oh...I think I see it now...but how can i demostrate if it is it's limit?

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.0but there is also the 1^n

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.0So it approaches 5/2 and 5/2 ?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

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

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.1No, 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.

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

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.1Seriously? An alternating sign in the exponent? No. No limit.

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.0I thought so, thanks :)
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