IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

toshi

lim 2-- (256 -- 7x)^1/8
x->0 -------------------------
(5x+32)^1/5 -- 2
the answer is 7/64.....plzz do it!

nishant

hi!well,on substituting 0,we get an indeterminate form( 0/0)...so apply,l'hsptl rule and get the answer..if u still don't get it,i will solve the question...

toshi

hey nishant i would b happy if u 'd solve the prob.

thanks!

nishant

fine...so,differentiating the above function,we get
lim x tends to 0 1/8(256-7x)^-7/8 .7divided by 1/5(5x+32)^-4/5 ..5 now substituting the limit,we get -1/8(2^8)^-7/8.7 divided by 1/5(2^5)-4/5..5 which gives -1/8 (2^-7)..7 divided by 1/5(2^-4)..5which on simpification gives answer as 7/64( as there is minus sign in numerator)..cheers..

vinu

Hey toshi,

the same prob has been given by "u" with subject as "problem related to continuity" on 19th feb,this year...

rajat

hi toshi,

i have with me a simpler solution than nishant's answer.

[ 2 -- (256 - 7x)^1/8 ] / [ (5x + 32)^1/5 -- 2 ]

in the numerator take 256 out , then it becomes [ 2-- 2(1--7x / 256)^1/8 ]

similarly ,in the denominator take 32 out .

then, expand both barackets using binomial formula and neglect higher powers of x .u'll get the answer.

by the way even i have one doubt .

how do u score nickels ??

toshi

yeah right vinu...but i was looking for a solution which matches wid the solution in tmh...! :D

toshi

well rajat u need to invite friends to this forum for gaining nickels nd if they register u get some extra nickels too.... ! :)

KAB

In TMH they have applied binomial expansion.When x is small so that x(square) can be neglected.
(1+x)(whole to the power n) is approximated to be 1+nx.

Ask & Discuss Questions with Community & Experts