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December 22, 2014

December 22, 2014

Posted by **Heather** on Wednesday, November 14, 2012 at 12:38pm.

- Calculus 2 -
**Count Iblis**, Wednesday, November 14, 2012 at 1:06pmf(x)= [1-1/(2x)]^(2 x)

Log[f(x)] = 2x Log[1-/(2x)] =

2 x [1/(2x) + O(1/x^2)] =

1 + O(1/x)

The limit for x to infinity of

Log[f(x)] is thus 1. The limit for x to infinity of f(x) is thus e, because the logarithmic function is continuous there.

- Calculus 2 -
**Steve**, Wednesday, November 14, 2012 at 2:01pmHmmm. Let u = 1/2x and you have

(1+u)^(-1/u) = 1 / (1+u)^(1/u)

as x->infinity, u->0, and we have

1/e

- Calculus 2 -
**Count Iblis**, Wednesday, November 14, 2012 at 2:11pmI see, I forgot a minus sign.

f(x)= [1-1/(2x)]^(2 x)

Log[f(x)] = 2x Log[1-/(2x)] =

2 x [-1/(2x) + O(1/x^2)] =

-1 + O(1/x)

So, you get exp(-1), using the fact that log is continuous there.

- Calculus 2 -
**Heather**, Wednesday, November 14, 2012 at 2:35pmthank you! we are supossed to use Lopital's rule though. Is there a different way to do it using Lopital's?

- Calculus 2 -
**Count Iblis**, Wednesday, November 14, 2012 at 3:43pmIf you have to use L'Hôpital, then that's close to how I did it above. After taking logarithms, you have:

Log[f(x)] = 2x Log[1-/(2x)]

You can write this as:

Log[1-1/(2x)] / (1/(2x))

This is then a 0/0 case. If you put

1/(2x) = u then this becomes:

Log[1-u]/u

and you need to compute the limit of u to zero.

- Calculus 2 -
**Heather**, Thursday, November 15, 2012 at 11:01amthank you so much!

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