So your question is how he can multiply infinite by infinite.... right?
consider infinite to be a some very large number.and if u multiply dis large no. wid another large number...then i will also be much much greater dat we cant detect.
That's right, my question is how he can claim a result of inf * inf when that is undefined
inf * inf is like something/0 *UNDEFINED*
Well same as what madrockz said but I think that he can do it but it just means that the end # you get will be undefined with it
I really doubt this to be the case but thanks alot for the input ! Any further opinion would be welcome !
@hartnn Maybe you have any ideas ?
Wait, what grade are you in...?
That would explain why I dont know it. Lol
i think the solution is incorrect x^(-1/2) = 0 when x->infinity so, denominator = 0*infinity = indeterminate
and we need one more iteration of L h rule 2x^(1/2) / ln x ----> inf/inf form
= 2 *(1/2) x^(-1/2) times x = x^(1/2) and x^(1/2) when x->infinity = infinity = final answer
hm hold on
eh, but wolf shows final answer = 0 i really can't make out where i made error....
there is a tricky chain rule of ln(ln(x) that I think we missed there
or no ?
d/dx (ln ln x) = 1/ (x ln x)
I go sort(x) ----> inf as well :O
oh wait! sorry, that was x^(1/2) and not x^(-1/2)
but if you differenciate it it will be minus
why would u diff it ?
as told earlier, in limits we consider inf*inf = inf
soo that's not undefined ? It actually is inf * inf = inf ?
only when x-> infinity without limits , its undefined
inf * - inf = - inf ?
seems madrocks was correct from the beginning... yes
but wait, how can I come to the final step in the first place ?
nothing nothinng I solved it ! The only thing that confused me was the inf * inf part, thanks !
I would like to ask some additional questions if possible, what if the limit is ----> -inf ? Would I operate the same way ? Are there any other undefined functions that would be "defined" under a specific situation in limits ? @hartnn
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