Question

evaluate limit x-> 0 (1+ cos^3 x)/ sin^2 x

Answer 1

you took a mistake here,
I think Question may be
Lim(x→0) ( 1-cos³x)/sin²x

Lim(x→0) ( 1 -cos³x )/sin²x

[ use, (a³ - b³) = (a - b)(a² + b² +ab)]

Lim(x→0) ( 1 - cosx)(1 + cos²x +cosx )/sin²x
[use, (1 - cosx ) = 2sin²x/2 ]

Lim(x→0) (2sin²x/2)( cos²x +cosx +1 )/sin²x

now, use ,
Lim{ f(x)→0} sinf(x)/f(x) = 1

2× Lim(x →0) {sinx/2/x/2}² × x²/4 ( cos²x + cosx + 1 )/{sinx/x}² × x²

2 × 1/4 × 3 = 3/2

hence, answer = 3/2