Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^{\sin \pi x}} \cr
& {\text{Evaluate }} \cr
& \mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^{\sin \pi x}} = {\left( {1 - 1} \right)^{\sin \pi }} = {0^0} \cr
& {\text{This limit has the form }}{{\text{0}}^0}.{\text{ }}\left( {{\text{Using theorem 2}}{\text{.12}}} \right) \cr
& {\text{Noting that }}{\left( {x - 1} \right)^{\sin \pi x}} = {e^{\sin \pi x\ln \left( {x - 1} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^{\sin \pi x}} = \mathop {\lim }\limits_{x \to 1} {e^{\sin \pi x\ln \left( {x - 1} \right)}} = {e^{\mathop {\lim }\limits_{x \to 1} \sin \pi x\ln \left( {x - 1} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to 1} \sin \pi x\ln \left( {x - 1} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\ln \left( {x - 1} \right)}}{{\frac{1}{{\sin \pi x}}}} = \frac{\infty }{\infty } \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left[ {\ln \left( {x - 1} \right)} \right]}}{{\frac{d}{{dx}}\left[ {\frac{1}{{\sin \pi x}}} \right]}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{{x - 1}}}}{{ - \pi \csc \pi x\cot \pi x}} \cr
& {\text{Evaluate}} \cr
& L = \frac{{\frac{1}{{1 - 1}}}}{{ - \pi \csc \pi \cot \pi }} = \frac{\infty }{\infty } \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = - \frac{1}{\pi }\mathop {\lim }\limits_{x \to 1} \frac{{{{\sin }^2}\pi x}}{{\left( {x - 1} \right)\cos \pi x}} \cr
& L = - \frac{1}{\pi }\mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left[ {{{\sin }^2}\pi x} \right]}}{{\frac{d}{{dx}}\left[ {\left( {x - 1} \right)\cos \pi x} \right]}} \cr
& L = - \frac{1}{\pi }\mathop {\lim }\limits_{x \to 1} \frac{{2\pi \sin \pi x\cos \pi x}}{{ - \pi \left( {x - 1} \right)\sin \pi x + \cos \pi x}} \cr
& {\text{Evaluating}} \cr
& L = - \frac{1}{\pi }\left( {\frac{{2\pi \sin \pi \cos \pi }}{{ - \pi \left( {1 - 1} \right)\sin \pi + \cos \pi }}} \right) \cr
& L = - \frac{1}{\pi }\left( {\frac{0}{{0 - 1}}} \right) = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^{\sin \pi x}} = \mathop {\lim }\limits_{x \to 1} {e^{\sin \pi x\ln \left( {x - 1} \right)}} = {e^{\mathop {\lim }\limits_{x \to 1} \sin \pi x\ln \left( {x - 1} \right)}} = {e^0} \cr
& \mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^{\sin \pi x}} = 1 \cr} $$
