Answer
$$\frac{1}{2}$$
Work Step by Step
Since
\begin{align*}
\lim _{r \rightarrow \infty} \frac{r+\sin r}{2 r+7-5 \sin r}&=\lim _{r \rightarrow \infty} \frac{1+\frac{\sin r}{r}}{2+\frac{7}{r}-\frac{5 \sin r}{r}}\\
&= \frac{\lim _{r \rightarrow \infty} (1)+\lim _{r \rightarrow \infty} \frac{\sin r}{r}}{\lim _{r \rightarrow \infty} (2)+\lim _{r \rightarrow \infty} \frac{7}{r}-\lim _{r \rightarrow \infty} \frac{5 \sin r}{r}}\\
&=\frac{1+0}{2+0-0}\\
&=\frac{1}{2}
\end{align*}
