Answer
$\frac{1}{4}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^3} + x - 2}} \cr
& {\text{Use the quotient property of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 1} \left[ {\sin \left( {x - 1} \right)} \right]}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^3} + x - 2} \right)}} \cr
& {\text{Evaluate the limit}} \cr
& = \frac{{\sin \left( {1 - 1} \right)}}{{{{\left( 1 \right)}^3} + \left( 1 \right) - 2}}{\text{ = }}\frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^3} + x - 2}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left[ {\sin \left( {x - 1} \right)} \right]}}{{\frac{d}{{dx}}\left[ {{x^3} + x - 2} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{\cos \left( {x - 1} \right)}}{{3{x^2} + 1}} \cr
& {\text{Evaluate the limit when }}x \to 1 \cr
& = \frac{{\cos \left( {1 - 1} \right)}}{{3{{\left( 1 \right)}^2} + 1}} \cr
& = \frac{1}{4} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^3} + x - 2}} = \frac{1}{4} \cr} $$
