62[(x+p)(x+q)(x+r)(x+s)]1/4 - x
([(x+p)(x+q)(x+r)(x+s)]1/4 - x)([(x+p)(x+q)(x+r)(x+s)]1/4 + x) /([(x+p)(x+q)(x+r)(x+s)]1/4 + x)
= ([(x+p)(x+q)(x+r)(x+s)]1/2 - x2)/([(x+p)(x+q)(x+r)(x+s)]1/4 + x)
= ([(x+p)(x+q)(x+r)(x+s)]1/2 - x2)([(x+p)(x+q)(x+r)(x+s)]1/2 + x2)/([(x+p)(x+q)(x+r)(x+s)]1/4 + x)([(x+p)(x+q)(x+r)(x+s)]1/2 + x2)
= ([(x+p)(x+q)(x+r)(x+s)] - x4)/([(x+p)(x+q)(x+r)(x+s)]1/4 + x)([(x+p)(x+q)(x+r)(x+s)]1/2 + x2)
now divide numerator and deno by x3
I think that will suffice :)
33Can't we approach like as x→∞
x+p≈x+q≈x+r≈x+s
so its GM which is equal to AM
therefore
given limit....
(x+p)+(x+q)+(x+r)+(x+s)
------------------------------------ - x
4
which becomes p+q+r+s
---------- ...............
4
Is this the answer?
62u did get the right answer i think...
but i have some reservations about this method..
basically bcos i am not sure if this approach will work for all problems
(x+p)+(x+q)+(x+r)+(x+s)
------------------------------------ - x
4
i think if u used this instead .. it wud be mroe convincing!
(1+p/x)+(1+q/x)+(1+r/x)+(1+s/x)
------------------------------------------------ - 1/x
4
But i am open to discussions if we could have some on this one.. i like the approach no doubt...
33i have tried this on two or three problems it worked for (x+p) type but not for one of these x*p or xp type(i don't remember which)...
62hmm.. as far as i can think.. i am not being able to make up some exception on this one right now...
it shoudl work fine.. but then i am only about 90% convinced... tht is y i said i have some reservations!
Though this is a very very good way to think about this limit