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Mathematics 13 Online
OpenStudy (osanseviero):

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

OpenStudy (osanseviero):

\[\left( -1^{n} \right)\left( \frac{ 5n+4 }{ 2n } \right)\]

OpenStudy (osanseviero):

Help, please?

OpenStudy (osanseviero):

:/

OpenStudy (tkhunny):

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.

OpenStudy (osanseviero):

oh...I think I see it now...but how can i demostrate if it is it's limit?

OpenStudy (osanseviero):

but there is also the -1^n

OpenStudy (osanseviero):

So it approaches 5/2 and -5/2 ?

OpenStudy (tkhunny):

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.

OpenStudy (osanseviero):

what I mean is that there is a (-1^n) multiplying all of that...so -5/2 is also a limit

OpenStudy (tkhunny):

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.

OpenStudy (osanseviero):

Oh...okk

OpenStudy (osanseviero):

So for this there isnt a limit, neither \[\frac{ 1 }{ 2 },2^{2}, \frac{ 1 }{ 2^{3} }\]

OpenStudy (tkhunny):

Seriously? An alternating sign in the exponent? No. No limit.

OpenStudy (osanseviero):

I thought so, thanks :)

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