Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {\frac{{dx}}{{\sqrt {2 - x} }}} \cr
& {\text{definition of improper integral}} \cr
& \int_{ - \infty }^0 {\frac{{dx}}{{\sqrt {2 - x} }}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{dx}}{{\sqrt {2 - x} }}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{{\left( {2 - x} \right)}^{ - 1/2}}} dx \cr
& {\text{evaluate the integral}} \cr
& = - \mathop {\lim }\limits_{a \to - \infty } \left. {\left( {\frac{{{{\left( {2 - x} \right)}^{1/2}}}}{{1/2}}} \right)} \right|_a^0 \cr
& {\text{simplify}} \cr
& = - 2\mathop {\lim }\limits_{a \to - \infty } \left. {\left( {\sqrt {2 - x} } \right)} \right|_a^0 \cr
& {\text{evaluate the limit}} \cr
& = - 2\left( {\sqrt {2 - 0} } \right) + 2\left( {\sqrt {2 - \left( { - \infty } \right)} } \right) \cr
& = - 2\sqrt 2 + \infty \cr
& = \infty \cr
& {\text{diverges}} \cr} $$