Question
The value of $$\mathop {\lim }\limits_{x \to 0} \left( {{{\left( {\sin \,x} \right)}^{\frac{1}{x}}} + {{\left( {1 + x} \right)}^{\sin \,x}}} \right),$$ where $$x>0$$ is :
A.
$$0$$
B.
$$-1$$
C.
$$1$$
D.
$$2$$
Answer :
$$1$$
Solution :
$$\eqalign{
& {\text{Given, }}\mathop {\lim }\limits_{x \to 0} \left( {{{\left( {\sin } \right)}^{\frac{1}{x}}} + {{\left( {1 + x} \right)}^{\sin \,x}}} \right) \cr
& = \mathop {\lim }\limits_{x \to 0} {\left( {\sin \,x} \right)^{\frac{1}{x}}} + \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\sin \,x}} \cr
& = {\left( {\sin \,0} \right)^{\frac{1}{0}}} + \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\sin \,x}} \cr
& = {\left( 0 \right)^\infty } + \mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\sin \,x}} \cr
& = 0 + {e^{\mathop {\lim }\limits_{x \to 0} \left[ {\log {{\left( {1 + x} \right)}^{\sin \,x}}} \right]}}\,\,\,\,\,\,\,\left( {\because \,{e^{\log \,x}} = a} \right) \cr
& = {e^{\mathop {\lim }\limits_{x \to 0} \left[ {\sin \,x\,\log \left( {1 + x} \right)} \right]}} \cr
& = {e^{\mathop {\lim }\limits_{x \to 0} \sin \,x \times \mathop {\lim }\limits_{x \to 0} \log \left( {1 + x} \right)}} \cr
& = {e^{\sin \left( 0 \right) \times \log \left( {1 + 0} \right)}} \cr
& = {e^{0 \times \log \left( 1 \right)}} \cr
& = {e^{0 \times 0}} \cr
& = {e^0} \cr
& = 1 \cr} $$