Problem 17 Short method Cos2x -1=2 cos ^2x - 1-1 = 2cos ^2x -2=2(cos^2x -1) =2(cosx +1)(cos x -1) Hence (cos2x -1)/(cos x -1)=2cos x +2 Hence Lt (X tends to zero) (2cosx +2) =2cos0+2=2*1+2=4 By direct substitution It is a short method .
Wo isliye bro kyuki Cos 0= 1 and 1-1=0 hoga So, 0/0 ye inderminant form hota hai (mtlb iska koi fixed answer, ya value nahi hota) Therefore we cannot cancel, cos 0-1/cos0-1
May see my solution. We cancel out from denominator and numerator ( cos x -1) Here X tends to zero ( not zero) Hence cos x is not cos 0 Hence cos x is not 1 Hence cos x -1 is not zero