IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ashish_90

_{[x ]}

_{[0 ]} [cos(tan x) - cos x] / x ⁴=

taruntanuj007

Use lHOspitals RUle its clearly 0/0 form!!!

lim (x->0) -sin(tanx)secx*secx - sinx / 4x^3

Now sin(tanx)/tanx = 1 and tanx / x = 1 thus sin(tanx)/x = 1

and now solve u can repeatedly differentiate to get a simpler limit or solve it.....

ashish_90

thats not the only way try another method

arvind.b

cos(tan x)-cos(x)/x^4

-2 sin(tan x + x /2) sin(tan x - x/2) / x^4

separate this into 2 limits

-2 sin(tanx + x/2) / x and sin(tan x -x /2 )/x^3

in both limits use the power series expansion of

tan x = x + x^3/3 + ....

and we get

(-2)(1)(1/6) = -1/3

I might have made a mistake of a sign, so please check.

avinash.sharma

arvind.b has given the correct guidance. Lets get the complete solution-

as CosA ? Cos B = 2Sin(A+B)/2 . Sin(B-A)/2

=  _{[ x][ 0]}  [2Sin(Tanx+x)/2 . Sin(x-Tanx)/2]/x⁴

we know that  _{[ x][ 0]} {(sinx)/x} = 1 so the above expression can be written as

= 2  _{[ x][ 0]}  [{Sin(Tanx+x)/2}/{(Tanx+x)/2}] . [{Sin{(x-Tanx)/2}/{(x-Tanx)/2}] . [{(Tanx+x)/2} . {(x-Tanx)/2}] . [1/x⁴]

factorizing x⁴ as x. x³

= (2/4) _{[ x][ 0]}   [{Sin(Tanx+x)/2}/{(Tanx+x)/2}] . _{[ x][ 0]}  [{Sin{(x-Tanx)/2}/{(x-Tanx)/2}] . _{[ x][ 0]}   {{(Tanx+x)/x} . _{[ x][ 0]}   {(x-Tanx)/x³}

 expanding Tanx = x + (x³/3) + (2x⁵/15)+ ...............

= (1/2) _{[ x][ 0]}   [{Sin(Tanx+x)/2}/{(Tanx+x)/2}] ._{[ x][ 0]}  [{Sin{(x-Tanx)/2}/{(x-Tanx)/2}] . _{[ x][ 0]}   {{(Tanx+x)/x} . _{[ x][ 0]}  [{x-( x + (x³/3) + (2x⁵/15)+ ....)}/x³]

 = (1/2) 1.1. (1+1). (-1/3)

=(1/2)(2)(-1/3)

= -(1/3)

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