Limits Continuity and Differentiability

JEE Main 2019 · 11 Jan, Evening shift · Q4273

MathematicsEasy179×since 2002Q4273

\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}

is equal to :

JEE Main 2019 · 11 Jan, Evening shift · Q4273

MathematicsEasy179×since 2002Q4273

\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}

is equal to :

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