Lt x->0 log(cos x)/ (sinx)²

Lt x->0 log [1 + (cosx - 1)]/(sinx)²                                         {Add and subtract 1}

Lt x->0 log [1 + (cosx - 1)](cosx - 1)/ (cosx - 1) (sin x)²          {Multiply divide by (cosx - 1)}

Lt x->0 {log [1 + (cosx - 1)]/ (cos x - 1)}  {(cosx - 1) / (sinx)²}

Lt x->0 {(log [1 + m])/ (m)} { - (1 - cosx)/ ??(sinx)²}

 Here m->0 as cosx - 1 ->0 when x-> o So, Standard limit Lt x->0  log[ 1 + x]/ x = 1

Lt x->0 {1} {- (1 - cosx)(x²) / (x²) (sinx)²                                            {Multiply divide by (sinx)²}

Lt x->0 - {(1- cosx) / (x²)} { (x²) / (sinx)²}

Using Standard limit Lt x->0 (1-cosx) / (x²) = 1/2 and Lt x->0  (x/ sinx)= 1, we get,

Lt x->0 - {1/2} { (1)² }

Lt x->0 -1/2

= -1/2

Anmol , Ur answer is indeed right . but I want to do it by a simple method called L Hopitals rule.

Any problem yielding 0/0 by direct substition can be solved by the approach.

This method coveys tht any  limit yielding 0/0

lim f(x)/g(x) = {d f(x) /dx } / {d g(x) /dx }

Here in the question by direct substitution we get 0/0 So apply L Hopitals rule

Lt (x->0) log(cos(x))/(sinx)² = Lt(x->0)d/dx {log(cos(x))} / d/dx {sin²x} = -1/2 cos²(x)  = -1/2 (By direct Substitution)