Originally Posted by Moo ---------------------------------- 2/ ---------------------------------- This one seems way too easy.... --Chris
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@ Mathstud, #15 : it looks complicated, huh ? Correct substitution, but what about this approximation : (Taylor series, that's what I think about when talking about approximations) when --> @ Chris : Yes
Originally Posted by Moo @ Mathstud, #15 : it looks complicated, huh ? Correct substitution, but what about this approximation : (Taylor series, that's what I think about when talking about approximations) when --> Dangit! That probably would have been easier...oh well...same result right?
Originally Posted by Moo 10/ Let So we get Now note that so So we have
Originally Posted by Mathstud28 Let So we get Now note that so So we have Ahahahaha...you beat me to it. I pretty much had what you had... --Chris
Originally Posted by Chris L T521 Ahahahaha...you beat me to it. I pretty much had what you had... --Chris Quick fingers
Originally Posted by Moo I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean?
Originally Posted by Mathstud28 I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean? a and b > 0 ? i don't know this relation, but i'd say no For the previous one, there's a simpler way, but i'll wait a while
Originally Posted by Moo a and b > 0 ? ?
Originally Posted by Mathstud28 ? Nothing mentioned about it, this means that your method won't work (I can see you dividing by a or b xD)
Originally Posted by Moo Nothing mentioned about it, this means that your method won't work (I can see you dividing by a or b xD) Yeah there goes that, and the Geometric Arithmetic mean thing is so nice. but here goes nothing Now for the second limt If we let We get So Hooray! That was cool! [tex]
Originally Posted by Mathstud28 I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean? I think something is missing as well...however, I think I almost have it...but its somewhat of a pain... --Chris EDIT: As usual...he beats me again...
Originally Posted by Chris L T521 I think something is missing as well...however, I think I almost have it...but its somewhat of a pain... --Chris EDIT: As usual...he beats me again... Haha...aww..now I feel bad ...sorry
Hi ! Originally Posted by Moo 11/ Let and since . As and ,
Originally Posted by flyingsquirrel Hi ! Let and since . As and , Awesome! But isn't this kind of using Cauchy's Mean Value Theorem...which is basically L'hopital's?
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