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limit n tend to infinity (n+3/n+1)^n the sequences is converge or diverge..? plz solve...

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armankhan | Student, Undergraduate | (Level 2) eNoter

Posted June 28, 2012 at 9:46 PM via web

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limit n tend to infinity (n+3/n+1)^n the sequences is converge or diverge..? plz solve the step by step

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted June 29, 2012 at 12:06 AM (Answer #1)

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First, recall that, for positive numbers,

`x = e^("ln" x)`



When you see a limit question of the form `1^(oo)`

(that is, `(n+3)/(n+1) -> 1`, `n-> oo` )

this is usually the way you want to solve it:



Let `Q = ( (n+3)/(n+1) )^n`

Then `Q = e^("ln" Q)`

`lim_(n-> oo) Q = e^(lim "ln" Q)`


So, what we want to start by figuring out, is:

`lim_(n->oo) "ln" Q`

or

`lim_(n->oo) "ln" ( ( (n+3)/(n+1) )^n)`






`"ln" ( ( (n+3)/(n+1) )^n) `

`= n "ln" ((n+3)/(n+1))`

`= ("ln" ((n+3)/(n+1)) )/ ((1)/(n))`

Now, as `n -> oo`, the numerator and the denominator both go to 0

Thus, we use l'Hopital's rule.



The derivative of

`"ln" ((n+3)/(n+1)) = "ln" (n+3) - "ln" (n+1)`

is

`(1)/(n+3) - (1)/(n+1) = ((n+1)-(n+3))/((n+1)(n+3)) = (-2)/((n+1)(n+3))`


The derivative of

`(1)/(n) = n^(-1)`

is

`-n^(-2)`



So, l'hopital's rule gives us that:

`lim_(n-> oo) ("ln" ((n+3)/(n+1)) )/ ((1)/(n))`

`= lim_(n-> oo) ((-2)/((n+1)(n+3)))/(-n^(-2))`

`= lim_(n-> oo) (2 n^2)/((n+1)(n+3))`

`= 2`




`lim_(n-> oo) Q = e^(lim "ln" Q) = e^(2)`


Thus the limit is: `e^(2)`

 

 

 

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