Posted by T.Gracken on August 29, 2002 at 18:57:14:
In Reply to: Re: questions??? posted by T.Gracken on August 29, 2002 at 18:51:16:
O.K. I screwed up on the fonts. this is a little better notation but still messed up. Anyway, just because I was trying to show off, doesn't mean the math is wrong. let me (or anyone here) know if what I write doesn't make sense and someone will try to further aid your quest for calculus knowledge.
: in your problem: : lim [as x -- > 0] of [1 - e-x]/[ex - 1]
: has the form 0/0 (that is, if you attempt to substitute 0 for x). that is one of the conditions necessary to use L'Hopital's rule.
: So, you can take the derivative of the numerator of the argument (the argument is the expression you are trying to figure the limit of) divided by the derivative of the denominator of the argument, and use this new expression.
: ...yeah, that's a little much. so, using your example: the argument is [1 - e-x]/[ex - 1]
: ...if you place 0 in for x (the limit value), we get 0/0. so use L'Hopitals rule to get
: derivative of numerator is e-x
: derivative of denominator is ex
: so now consider lim [as x -- > 0] of [e-x]/[ex]
: which is easily evaluated as 1. (cause now we can just plug 0 in and get a real number)
: hope that helps. if not...
:
: ...tell me.