Answer
$0$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{{e^{ - x}}}}{{\left( {\pi /2} \right) - {{\tan }^{ - 1}}x}} \cr
& {\text{Use the quotient property of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left[ {{e^{ - x}}} \right]}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\left( {\pi /2} \right) - {{\tan }^{ - 1}}x} \right)}} \cr
& = \frac{{{e^{\mathop {\lim }\limits_{x \to \infty } \left( { - x} \right)}}}}{{\frac{\pi }{2} - \mathop {\lim }\limits_{x \to \infty } \left( {{{\tan }^{ - 1}}x} \right)}} \cr
& = \frac{0}{{\frac{\pi }{2} - \frac{\pi }{2}}} \cr
& = \frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{e^{ - x}}}}{{\left( {\pi /2} \right) - {{\tan }^{ - 1}}x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {{e^{ - x}}} \right]}}{{\frac{d}{{dx}}\left[ {\left( {\pi /2} \right) - {{\tan }^{ - 1}}x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{ - {e^{ - x}}}}{{0 - \frac{1}{{1 + {x^2}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{ - {e^{ - x}}}}{{\frac{{{x^2} + 1 - 1}}{{1 + {x^2}}}}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{ - \left( {1 + {x^2}} \right){e^{ - x}}}}{{{x^2}}} \cr
& = \frac{{ - \mathop {\lim }\limits_{x \to \infty } \left( {1 + {x^2}} \right)\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - x}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {{x^2}} \right)}} \cr
& = -\lim_{x \to \infty } \left(1 + \frac{1}{x^2} \right)\lim_{x \to \infty } \left(e^{ - x} \right) \cr
& \text{As the limit is in the form }1\cdot 0\text{ therefore} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{{e^{ - x}}}}{{\left( {\pi /2} \right) - {{\tan }^{ - 1}}x}} = 0 \cr} $$