Question
$$\mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} {\left[ {1 + {{\left( {\cos \,x} \right)}^{\cos \,x}}} \right]^2}$$ is equal to :
A.
does not exist
B.
$$1$$
C.
$$e$$
D.
$$4$$
Answer :
$$4$$
Solution :
$$\eqalign{
& {\text{Given }}\mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} {\left[ {1 + {{\left( {\cos \,x} \right)}^{\cos \,x}}} \right]^2} \cr
& {\text{Let }}y = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} {\left( {\cos \,x} \right)^{\cos \,x}} \cr
& \log \left( y \right) = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \left( {\cos \,x} \right)\log \,\cos \,x \cr
& \log \left( y \right) = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \frac{{\log \left( {\cos \,x} \right)}}{{\sec \,\left( x \right)}}\,\,\,\left( {\frac{\infty }{\infty }{\text{ form}}} \right) \cr
& {\text{Applying L'Hospital's rule}} \cr
& \log \left( y \right) = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \frac{{ - \sin \,x}}{{\cos \,x\left( {\sec \,x\,\tan \,x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to {{\frac{\pi }{2}}^ - }} \left( { - \cos \,x} \right) \cr
& = 0 \cr
& \therefore \,\,y = {e^0} = 1 \cr
& {\text{Now, limit is}}\,{\left( {1 + 1} \right)^2} = {2^2} = 4 \cr} $$