Question
$$\mathop {\lim }\limits_{x \to 0} {\left\{ {\frac{{1 + \tan \,x}}{{1 + \sin \,x}}} \right\}^{{\text{cosec}}\,x}}$$ is equal to :
A.
$$\frac{1}{e}$$
B.
$$1$$
C.
$$e$$
D.
$${e^2}$$
Answer :
$$1$$
Solution :
$$\eqalign{
& {\text{Consider }}\mathop {\lim }\limits_{x \to 0} {\left\{ {\frac{{1 + \tan \,x}}{{1 + \sin \,x}}} \right\}^{{\text{cosec}}\,x}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left[ {{{\left( {1 + \frac{{\sin \,x}}{{\cos \,x}}} \right)}^{\frac{{\cos \,x}}{{\sin \,x}}}}} \right]}^{\frac{1}{{\cos \,x}}}}}}{{{{\left( {1 + \sin \,x} \right)}^{\frac{1}{{\sin \,x}}}}}} \cr
& {\text{We know, }}\mathop {\lim }\limits_{n \to 0} {\left( {1 + \frac{1}{n}} \right)^n} = e \cr
& \therefore \,\mathop {\lim }\limits_{x \to 0} \frac{{{{\left[ {{{\left( {1 + \frac{{\sin \,x}}{{\cos \,x}}} \right)}^{\frac{{\cos \,x}}{{\sin \,x}}}}} \right]}^{\frac{1}{{\cos \,x}}}}}}{{{{\left( {1 + \sin \,x} \right)}^{\frac{1}{{\sin \,x}}}}}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left[ {{{\left( {1 + \frac{1}{{\frac{{\cos \,x}}{{\sin \,x}}}}} \right)}^{\frac{{\cos \,x}}{{\sin \,x}}}}} \right]}^{\frac{1}{{\cos \,x}}}}}}{{\left[ {{{\left( {1 + \frac{1}{{{\text{cosec}}\,x}}} \right)}^{{\text{cosec}}\,x}}} \right]}} \cr
& = \frac{{{e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos \,x}}}}}}{e} \cr
& = \frac{e}{e} \cr
& = 1 \cr} $$