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

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

tkhunny
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0but there is also the 1^n

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

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

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