JEE MAIN - Mathematics (2015 (Offline) - No. 3)
$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$$ is equal to
2
$${1 \over 2}$$
4
3
Explanation
Multiply and divide by $$x$$ in the given expression, we get
$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x^2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan \,4x}}$$
$$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$$
$$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$$
$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x^2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan \,4x}}$$
$$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$$
$$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$$
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