Question

Lim x->0 (log(cosx+sinx))/x=????

Answer 1

Answer:

-1/2

Step-by-step explanation:

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

Answer 2

Step-by-step explanation:

See the picture.Thank you.