Verified Solution • 2025–26 Session moderate QJEEM13Ma4d12: JEE Main 2013 Mathematics - Class 12 JEEM13Ma4d12 Solution limx→0(1−cos2x)(3+cosx)xtan4x\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}x→0limxtan4x(1−cos2x)(3+cosx) is equal to Multiple Choice Options A −14- {1 \over 4}−41 B 12{1 \over 2}21 C 1 D 2 Step-by-Step Verified Solution Multiply and divide by xxx in the given expression, we get limx→0(1−cos2x)x2(3+cosx)1.xtan 4x\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x_2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}x→0limx2(1−cos2x)1(3+cosx).tan4xx =limx→02sin2xx2.3+cosx1.xtan4x= \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan 4x}}=x→0limx22sin2x.13+cosx.tan4xx =2limx→0sin2xx2.limx→03+cosx.limx→0xtan4x= 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}=2x→0limx2sin2x.x→0lim3+cosx.x→0limtan4xx =2.414limx→04xtan4x=2.4.14=2= 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2=2.441x→0limtan4x4x=2.4.41=2 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring