## Calculus, 10th Edition (Anton)

$$- 1$$
\eqalign{ & \mathop {\lim }\limits_{x \to {\pi ^ + }} \frac{{\sin x}}{{x - \pi }} \cr & {\text{Evaluate using theorem 1}}{\text{.2}}{\text{.2 }}\left( {{\text{law }}\left( d \right)} \right) \cr & = \frac{{\mathop {\lim }\limits_{x \to {\pi ^ + }} \left( {\sin x} \right)}}{{\mathop {\lim }\limits_{x \to {\pi ^ + }} \left( {x - \pi } \right)}} \cr & {\text{Using law }}\left( a \right) \cr & = \frac{{\sin \pi }}{{\pi - 0}} \cr & = \frac{0}{0} \cr & {\text{The numerator and denominator have a limit of 0}}{\text{, so the limit is }} \cr & {\text{an indeterminate form of type 0/0}}{\text{. Use L'Hopital's rule}} \cr & \mathop {\lim }\limits_{x \to {\pi ^ + }} \frac{{\sin x}}{{x - \pi }} = \mathop {\lim }\limits_{x \to 0} \frac{{d/dx\left( {\sin x} \right)}}{{d/dx\left( {x - \pi } \right)}} \cr & = \mathop {\lim }\limits_{x \to {\pi ^ + }} \frac{{\cos x}}{1} \cr & {\text{evaluating}} \cr & = \cos \pi \cr & = - 1 \cr & {\text{then }}\mathop {\lim }\limits_{x \to {\pi ^ + }} \frac{{\sin x}}{{x - \pi }} = - 1 \cr}