```                   plz solve -
Lt(x->0) (log(cos x)/(sinx)^2)
```

2 years ago

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```                    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
```
2 years ago
```                    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)
```
2 years ago