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Determine lim as n approaches infinity (1+(1/n))^n

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hellome234 | Salutatorian

Posted March 5, 2012 at 9:45 AM via web

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Determine lim as n approaches infinity (1+(1/n))^n

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 5, 2012 at 10:49 AM (Answer #1)

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The limit `lim_(n->oo)(1+ 1/n)^n` has to be determined.

The limit can be determined using synthetic proof which is one that begins with statements that are already known and progresses in steps until the required statement is proved.

It is known that `ln x = int_(1)^t (1/t) dt`

For t lying in any interval defined as [1, 1+1/n]

`1+1/n <= 1/t <= 1`

Take the following definite integral of the three sides.

`int_(1)^(1+ 1/n) 1+1/n dt <= int_(1)^(1 + 1/n) (1/t) dt <= int_(1)^(1 + 1/n) 1 dt`

=> `1/(1+n) <= ln(1 + 1/n) <= 1/n`

=> `e^(1/(1+n)) <= 1 + 1/n <= e^(1/n)`

take the (n+1)st power of the left

=> `e <= (1 + 1/n)^(n+1)`

and take the nth power of the right

=> `(1+1/n)^n <= e`

These give `(1+1/n)^n <= e <= (1 +1/n)^(n+1)`

Dividing `e <= (1 +1/n)^(n+1)` by `(1 + 1/n)` gives

`e/(1+1/n) <= (1+ 1/n)^n`

=> `e/(1+1/n) <= (1+ 1/n)^n <= e`

Using the squeeze theorem this gives `(1+1/n)^n` tends to e as n tends to infinity.

The value of `lim_(n->oo)(1+ 1/n)^n = e` .

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