IIT JEE , AIEEE Forum - Mathematics , Trignometry

ayshwarya

iamnt criticising anybody ok d problm is

2 2

Lim [1/x - 1/Sin x ] =? if donot know ill send d soln

x->0

spideyunlimited

is the answer 0?

ayshwarya

no d ans is -1/3 pk how do u get 0 can u explain ur medod it may b useful 2 all our frnds in many quesns

spideyunlimited

when x -> 0 , sin x also tends to 0.

raulrag009

yeah

i'm gettin -1/3

by l-hospital rule

ayshwarya

can u explain d soln

sboosy

lim(x-0) [1/x-1/sinx]
=(sinx-x)/xsinx
consider sinx-x/x cubed
by applying l'hospital's rule we get value to b -1/6(when x tends to 0)
so given = (sinx-x)/x cubed * x square /sinx
so answer is 0

sboosy

sorry for being hasty
i didnt notice the square in the question
next time type the word square if u dont have special editor
anyway using same technique
answer is
-1/6*(1+1)
=-1/3

raulrag009

u will hav to differentiate it 5 times to get the answer

take lcm

u will get

(sin^2x -x^2)/x^2sin^2x

differntiate it

1 derivative will be (sin2x-2x)/2xsin^2x + x^2sin2x

As it is still 0/0 form

differntiate it

At the end after about 4 to 5 differentiations

u will get

-8cos2x/20cos2x+ 4 cosx

{i hav eliminated terms which result in 0 }

Now

As cos 0 =1

therefore ans is -8/24 =-1/3

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