## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{r\to\infty}\frac{r+\sin r}{2r+7-5\sin r}=\frac{1}{2}$$
$$A=\lim_{r\to\infty}\frac{r+\sin r}{2r+7-5\sin r}$$ Here we can divide both numerator and denominator by $r$, the highest power of $r$ in the denominator: $$A=\lim_{r\to\infty}\frac{1+\frac{\sin r}{r}}{2+\frac{7}{r}-\frac{5\sin r}{r}}$$ - We already know $\lim_{r\to\infty}(7/r)=0$ - For $\lim_{r\to\infty}(\sin r)/r$: As $-1\le\sin r\le1$, we have $(-1/r)\le(\sin r/r)\le(1/r)$. But we know that $\lim_{r\to\infty}(-1/r)=\lim_{r\to\infty}(1/r)=0$ Therefore, following Squeeze Theorem: $\lim_{r\to\infty}(\sin r)/r=0$ That means, $$A=\frac{1+0}{2+0-5\times0}=\frac{1}{2}$$