NISHANT SIR, PL. XPLAIN

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.....

10 Answers

62
Lokesh Verma ·

see 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
tapanmast Vora ·

OH OK....

YEAH.... I WAS THINKIN HOW CUD V GET BY DIVIDING BY X3????
NOW ITS FINE....

SORT OF ANTI CLIMAX..... LOL....

62
Lokesh Verma ·

lol... okie :)

11
Anirudh Narayanan ·

What do u mean by dirty problem, bhaiya? [3][3][3][3][3][3][3]

1
skygirl ·

this is possible wid AM-GM as well... isnt it//

33
Abhishek Priyam ·

yes...

1
spiderman ·

i want the solution by am gm if possible ! pls

21
tapanmast Vora ·

spidy
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.....

62
Lokesh Verma ·

as 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 :)

33
Abhishek Priyam ·

heheehehe...

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