SOLUTION 28: a.) Start with $$\displaystyle{ -1 \le \sin x \le +1 } ,$$ so that $$\displaystyle{ 3x-1 \le 3x+\sin x \le 3x+1 }$$ and (Assume that $x>0$.) $$\displaystyle{ { 3x-1 \over 2x } \le { 3x+\sin x \over 2x } \le { 3x+1 \over 2x } } .$$ Then $$\displaystyle{ \lim_{x \to \infty} \ { 3x-1 \over 2x } } = \displaystyle{  \infty "\over \infty } = \displaystyle{ \lim_{x \to \infty} \ { 3x-1 \over 2x } \cdot { 1/x \over 1/x } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { 3x/x -1/x \over 2x/x } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { 3x/x -1/x \over 2x/x } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { 3 - \frac{1"}{\infty} \over 2 } }$$ $$= \displaystyle{ 3 - 0 \over 2 }$$ $$= \displaystyle{ 3 \over 2 } .$$ In a similar fashion it can easily be shown that $$\displaystyle{ \lim_{x \to \infty} \ { 3x+1 \over 2x } } = \displaystyle{ 3 \over 2 } .$$ Since $$\displaystyle{ \lim_{x \to \infty} \ { 3x-1 \over 2x } } = \displaystyle{ 3 \over 2 } = \displaystyle{ \lim_{x \to \infty} \ { 3x+1 \over 2x } } ,$$ it follows from the Squeeze Principle that $$\displaystyle{ \lim_{x \to \infty} \ { 3x+ \sin x \over 2x } } = \displaystyle{ 3 \over 2 } .$$

SOLUTION 28: b.) $$\displaystyle{ \lim_{x \to \infty} \ { 3x+ \sin x \over 2x } } = \displaystyle{  \ 3 \cdot \infty + (some \ number \ between \ -1 \ and \ +1) \ " \over 2 \cdot \infty } = \displaystyle{  \ \infty + (some \ number \ between \ -1 \ and \ +1) \ " \over \infty } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule.) $$= \displaystyle{ \lim_{x \to \infty} \ { 3+ \cos x \over 2 } } ,$$ which DOES NOT EXIST since $$\displaystyle{ -1 \le \cos x \le +1 }$$

SOLUTION 28: c.) The answers to parts a.) and b.) tell us that l'Hopital's Rule may give us a wrong answer if the answer is  does not exist." We can only be sure that l'Hopital's Rule gives us the correct answer if the answer is finite, $+ \infty$, or $- \infty$ .