IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sagar90

_{[ x]}

_{[ 0]} tanx-sinx/sin3x-3sinx

aditya_arora04

Hi,

Look,

tanx-sinx/sin3x-3sinx

sinx(secx-1)/sin3x-3sinx

divide num and denom by x. then, do further manipulation.

(sinx/x)(secx-1)/(3sin3x/3x - 3sinx/x)

put the limits to get the answer

sagar90

it will come zero yaar

aditya_arora04

Look, its answer is 0, but there is a mistake in sol.

correction, write sin3x = 4sin³x - 3sinx, take out sinx as common, then put x=0,

answer is 0.

sagar90

no the answer is not zero its something 1/8

sagar90

aditya please try the 3rd question posted by me

nunknown91

see
on applying 'L' hospital
_{[x ]}

_{[ 0]} (sec² x - cosx)/(12sin² x . cosx)

= _{[x ]}

_{[0 ]} (1-cos³ x)/(3sin² 2x . cosx)

=_{[x ]}

_{[0 ]} (1-cos³ x)/(3sin2x.sin2x)

Also we have a formula that

for the following set

{ x, sinx, sin^-1 x, tanx, tan^-1 x, sinhx, tanhx, sinh^-1 x, tanh^-1 x }

following formula holds good

_{[x ]}

_{[0 ]} [1 - cos^p (ax)]/f(mx) . g(nx) = (p . a² )/2mn

Apply it and u will get it as 1/8

rate me if it was useful

computer001

more simple::

(tanx-sinx)/4sin^3_x

_{^{(write sin3x as 3sinx -4sin^3x)}}

_{^{cancel 1 sinx}}

^{_{u will get 1-cosx/4coxsin}2}x

write sin^2x as 1-cos^2x=(1-cosx)(1+cosx)

then we get a proper form in which we can put x=0

nunknown91

but the ans is 1/8

computer001

in my method it comes out 1/8 too...i checked it up.. :|

sagar90

thank u nuknown91 and computer001
i got it

raulrag009

yes it is 1/8

sagar90

guys i think it will come -1/8
bcoz we forgot to take d -ve sign of -4sin^3x

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