Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 660 off


plz solve -

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

thanx in advance..

4 years ago


Answers : (2)


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

4 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)

4 years ago

Post Your Answer

if a 2, b 2, c 2 are in A.P. then prove that.. a/b+c, b/c+a, c/a+b
If a 2 , b 2 , c 2 are in A.P., so are a 2 +ab+bc+ca, b 2 +ab+bc+ca, and c 2 +ab+bc+ca i.e. (a+b)(c+a), (b+c)(a+b), (c+a)(b+c) are in A.P. Multiplying the three terms by (a+b+c)/...
mycroft holmes 7 months ago
If a 2 , b 2 , c 2 are in A.P., so are a 2 +ab+bc+ca, b 2 +ab+bc+ca, and c 2 +ab+bc+ca i.e. (a+b)(c+a), (b+c)(a+b), (c+a)(b+c) are in A.P. Multiplying the three terms by (a+b+c)/...
mycroft holmes 7 months ago
it is very easy use b 2 -a 2 =c 2 -b 2
dinesh 8 months ago
Number of positive integral solutions to equation x+y+z+w=10 (x>=0,y>=0,z>=2,w>=2) is PlZZZZZZZZZZ answer soon.
But the answer given is 84!!!!!!!!!.How????????????????.Plz provide the answer in a complete manner.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
gandhi1234 19 days ago
But z>=2 and w>=2 and you take z>=1 and w>=1.How is it possible?My answer is coming out to be 7C3.Is it correct???????????
gandhi1234 18 days ago
To Admin when i post my answer it did not take My answer and show a message answer already exists.plz help me. anywhy answer is
jagdish singh singh 19 days ago
explain :|z-4|3
Hii the question is incomplete .Kindly repost the question and we will get back at you
Sourabh Singh 22 days ago
Dear student, The question is incomplete. Pls post correct question again Thanks
Nishant Vora 22 days ago
range of sin^8x+cos^8x
HInt: Use a 2 + b 2 = (a+b) 2 – 2ab twice to reduce the power. Then try to make the equation in terms of sin2x only. Then you can easily find its range. Thanks.
Vijay Mukati 9 months ago
range is not [0,infinity]
Prajwal Kavad 9 months ago
If a,b,c be the unit vectors such that b is not parallal to c and a X(2bXc)=b then the angle that a makes with b and c are respectively where (X represents cross product)
We need to look at only one relation Taking dot product both sides wrt a the LHS will become zero (obviously) hence also now we can expand the given vector triple product as as a.b=0 and...
Riddhish Bhalodia 2 months ago
View all Questions »

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 4,500 off

Get extra R 660 off

More Questions On Algebra

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!!
Click Here for details