Posted by mskip on August 29, 2002 at 19:03:03:
In Reply to: Re: questions??? posted by T.Gracken on August 29, 2002 at 18:51:16:
: : : : : I am trying to solve this limit:
: : : : : lim 1 - e^-x / e^x - 1
: : : : : as X aproaches 0
: : : : : your help will be greatly appreciated thanx
: : : : : Michael
: : : : first... if it is simple, then why are you asking?
: : : : (o.k. that probably wans't called for, but I love sarcasm...)
: : : : second... have you seen L'Hopital's rule?
: : : : (not sarcastic... but my answer depends on how far you have delved into the subject)
: : : lol!!! Well I guessed that this would be simple for some of the genuises unlike me...lol!!
: : : I have heard L'Hopitals but maybe you can refamiliarize me. I can probably remember if you remind me.
: : : Thanx Michael
: : Okay now that I have done some research from my text it seems that I need to remember how to do integration
: : uhhhhh, lim f(x) / g(x) =
: : as X appraoches c
: : f'(x) / g'(x)
: : so do I intergrate first with the 0 in the formula or without?
: : Michael
: first... (i seem to start that way more than i should, but) thanks for having a sense of humor and not getting all uptight.
: second... (cause that usually comes after first), L'Hopital's rule lets you use derivatives (in some cases) to figure out (sometimes) painful limits.
: 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.
I really thank youfor this one. I understand how you got that, but still not totally clear about the derivative. I have forgotten how to do derivativative, I thought you had to multiply by the exponent and (n-1 in the exponent) right? Or is this integration
How did you get 1 - e^-x as e^-x and e^x -1 as e^x
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