I have this two problem to resolve with the ε-δ deﬁnition of a limit of succession lim n → ∞ 2 n + 3 3 n − 50 = 2 3

Question

I have this two problem to resolve with the ε-δ deﬁnition of a limit of succession      lim          n      →      ∞                  2      n      +      3              3      n      −      50        =      2    3

Mollie Roberts · Accepted Answer

For the first one, you have to prove that, given      ε  &amp;gt;  0  ,     ∃     N  ∈      N      such that         |                  2        n        +        3                    3        n        −        50              −          2      3        |       &amp;lt;  ε    ∀     n  ≥  N  .So consider the fact that            2      n      +      3              3      n      −      50        −      2    3    =            6      n      +      9              9      n      −      150        −      (                  6        n        −        100                    9        n        −        150              )    =      109          9      n      −      150           &amp;lt;      109    n      ∀  n  ≥  19.Therefore, for all      n  ≥  19  ,     0  &amp;lt;      |                  2        n        +        3                    3        n        −        50              −          2      3        |    =            2      n      +      3              3      n      −      50        −      2    3    =      109          9      n      −      150        &amp;lt;      109    n    ,and you should be able to use the Archimedean property of real numbers to complete the      ε  −  n    proof.For the second one, you must show that given      x  ∈      R    ,     ∃     N  ∈      N      such that               3              n        2            −      12      n      +      1              n      +      25        &amp;gt;  x    ∀  n  ≥  N  .   To do this, just use polynomial long division and then show that you can find an N that does this for each x

I have this two problem to resolve with the ε-δ deﬁnition of a limit of succession <munder>

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Answer & Explanation

New Questions in Calculus 1