function

sin (sin^-1x ) will be x only
same for cos and tan
so there is nothing difficult here ( or m i missing something ? )
for S_n, t_r=x^3r-2, t_r+1=x^3r+1

t_r+1=x³t_r

so S_n=x+x⁴+x⁷+x¹⁰+..... =x(1-x⁶ⁿ)1-x³

for C_n, similarly t_r+1=x³t_r
so C_n=x²+x⁵+x⁸+x¹¹+..... =x²(1-x⁶ⁿ)1-x³

similarly

T_n= x³(1-x⁶ⁿ)1-x³
now since x ε (0 , Π/4)
i.e x > 0 and x < 1
so x > x² > x ³, so S_n > C_n > T_n

so S_n + C_n + T_n = (x+x²+x³)(1-x⁶ⁿ)1-x³ = x(1+x+x²)(1-x⁶ⁿ)1-x³ =x(1-x⁶ⁿ)1-x

so lim n ->∞ S_n + C_n + T_n = x1-x

hey yes....i myself missed out the interval of x... thanxx..

shouldnt da limit be zero instead of ∞? the standard limit that qwerty has used applies for n→0 and nt ∞.

@ aditya i applied n→∞ only

fraction tends to zero when multiplied to itself infinite times