Question 300 of 401MediumMCQ
\[\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}\] is equal to
Solutions · 1
Official solution
Multiply and divide by x in the given expression, we get
\[\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x^2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}\]
\[ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan \,4x}}\]
\[ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}\]
\[ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2\] Answer the question to unlock