Answer
$$ - \pi $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {\pi x} \right)}}{{x - 1}} \cr
& {\text{Let }}t = \pi x - \pi \cr
& x \to 1 \Rightarrow t = \pi - \pi = 0 \cr
& t \to 0{\text{ when }}x \to 1 \cr
& and \cr
& x = \frac{t}{\pi } + 1 \cr
& {\text{Substituting}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {\pi x} \right)}}{{x - 1}} = \mathop {\lim }\limits_{t \to 0} \frac{{\sin \left( {t + \pi } \right)}}{{\frac{t}{\pi } + 1 - 1}} \cr
& = \mathop {\lim }\limits_{t \to 0} \frac{{\sin \left( {t + \pi } \right)}}{{\frac{t}{\pi }}} \cr
& = \pi \mathop {\lim }\limits_{t \to 0} \frac{{\sin \left( {t + \pi } \right)}}{t} \cr
& {\text{Use the identity }}\sin \left( {\theta + \pi } \right) = - \sin \theta .{\text{ we have}} \cr
& = - \pi \mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t} \cr
& {\text{Evaluating the limit}} \cr
& = - \pi \left( 1 \right) \cr
& = - \pi \cr} $$
