calculus question

[gusl]

how would you compute ?
This is relating to a common fallacy about probabilities.

Comments

[puellavulnerata.livejournal.com]

Clearly, since n does not appear in the expression, it's 1-(x-1\over x)^x. :)

[gustavolacerda.livejournal.com]

fixed. thx. :-)

[puellavulnerata.livejournal.com]

Better. :)

Anyway, with an application of l'Hopital's rule it's easy to show that the limit is 1-1/e.

[gustavolacerda.livejournal.com]

did you use pencil and paper?

[gustavolacerda.livejournal.com]

what are the f and g? http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

[puellavulnerata.livejournal.com]

Yeah. Move the 1- outside the limit, so you're looking for ((n-1)/n)^n as n approaches infinity. Then take logarithms to get an indeterminate form with zero times infinity inside the limit.

[gustavolacerda.livejournal.com]

I have: ((n-1)/n)^n = exp{n log((n-1)/n)}

as n->infty, the n->infty and the log goes to 0... but I can't go further. I should study http://en.wikipedia.org/wiki/Indeterminate_form#Evaluating_indeterminate_forms

[puellavulnerata.livejournal.com]

Yeah. Now you pull the exp outside the limit, and rewrite it as n/(1/log(...)), and you have infinity over infinity.

[gustavolacerda.livejournal.com]

thanks.

[jcreed.livejournal.com]

I got it by remembering the "compound interest" definition of the exponential

and manipulating the thing you gave until it resembled that enough.

[jcreed.livejournal.com]

To be more specific: pull the 1- outside the limit, and then notice that (n-1)/n is the same thing as 1 - (1/n). This means my x in this case is -1, so I get 1-e^(-1).

My brain-registers fill up pretty fast, and even then I didn't feel the need for pen and paper for this way. I think had I done L'Hopital's I would have.

[rdore.livejournal.com]

ditto.