21OH OK....
YEAH.... I WAS THINKIN HOW CUD V GET BY DIVIDING BY X3????
NOW ITS FINE....
SORT OF ANTI CLIMAX..... LOL....
Q3 lim x->infinity [(x+p)(x+q)(x+r)(x+s)]^1/4 - x x SOLN. ([(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
this was a soln by NIshant Sir,
Sir I didnt get how can we get da ans by dividing num and den by x3 ????
can u pl. give da complete soln.....
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10 Answers
62see this is a dirty problem...
Nothing very intellectually great about it
it comes from this...
{(x4+ax3+bx2+cx+d)1/4-x}
can be written as
x{(1+a/x+b/x2+c/x3+d/x4)1/4-1}
using the taylor series expansion for this, you will get that (1+a/x+b/x2+c/x3+d/x4)1/4 is approx 1+a/4
from then on it is simple... in the above question, a=(p+q+r+s)
21
11What do u mean by dirty problem, bhaiya? [3][3][3][3][3][3][3]
21spidy
COPYRITED BY ABHISHEK....... LOL............
as x tends to infinty, x+p = x+q = x+r = x+s;
therefor GM = AM;
so questn bcums {4x + p + q + r + s}/4 - x;
so u get da ans.....
62as x tends to infinty, x+p = x+q = x+r = x+s;
as i said is not true.. but the spirit of his proof is awesome.
I think you can better use
as x tends to infinty, 1+p/x = 1+q/x = 1+r/x = 1+s/x
now you can use AM GM inequality.
But the credit goes to abhishek alone :)