The value of \mathop {\lim }\limits_{h o 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} ight) - \cos \left( {{\pi \over 6} + h} ight)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } ight)}}} ight\} is :

Question

The value of \mathop {\lim }\limits_{h 	o 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} ight) - \cos \left( {{\pi \over 6} + h} ight)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } ight)}}} ight\} is :

Accepted Answer

Answer: A. {4 \over 3} Let L = \mathop {\lim }\limits_{h 	o 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} ight) - \cos \left( {{\pi \over 6} + h} ight)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } ight)}}} ight\} \Rightarrow L = \mathop {\lim }\limits_{h 	o 0} 2 	imes 2\left\{ {{{{{\sqrt 3 } \over 2}\sin \left( {{\pi \over 6} + h} ight) - {1 \over 2}\cos \left( {{\pi \over 6} + h} ight)} \over {2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } ight)}}} ight\} \Rightarrow L = \mathop {\lim }\limits_{h 	o 0} 2 	imes 2\left\{ {{{\cos {\pi \over 6}\sin \left( {{\pi \over 6} + h} ight) - \sin {\pi \over 6}\cos \left( {{\pi \over 6} + h} ight)} \over {2\sqrt 3 h\left( {\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh } ight)}}} ight\} \Rightarrow L = \mathop {\lim }\limits_{h 	o 0} 2 	imes 2\left\{ {{{\sin \left( {{\pi \over 6} + h - {\pi \over 6}} ight)} \over {2\sqrt 3 h\cos \left( {h + {\pi \over 6}} ight)}}} ight\} \Rightarrow L = {4 \over {2\sqrt 3 }}\mathop {\lim }\limits_{h 	o 0} \left\{ {{{\sin \left( h ight)} \over {h\cos \left( {h + {\pi \over 6}} ight)}}} ight\} \Rightarrow L = {4 \over {2\sqrt 3 }} 	imes {2 \over {\sqrt 3 }} = {4 \over 3}

The value of $$\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$$ is :

Answer

Explanation

More JEE Math questions