4.49 Let \( S=\left\{\left.\frac{2 x+3}{x+2} \right\rvert\, x \geqslant 0\right\} \). If there is a smallest number \( M \) such that all elements of \( S \) are not greater than \( M \), then \( M \) is the least upper bound (supremum) of \( S \). If there is a largest number \( m \) such that all elements of \( S \) are not less than \( m \), then \( m \) is the greatest lower bound (infimum) of \( S \), then
(A) \( m \) is in \( S \), \( M \) is not in \( S \).
(B) \( M \) is in \( S \), \( m \) is not in \( S \).
(C) \( m \) and \( M \) are both in \( S \).
(D) \( m \) and \( M \) are both not in \( S \).
(E) \( M \) does not exist.
(12th American High School Mathematics Examination, 1961)
[Solution] Let $y=\frac{2 x+3}{x+2}$, then $y=2-\frac{1}{x+2}$. Therefore, when $x \geqslant 0$, $y$ is an increasing function of $x$.
When $x=0$, $y_{\text {min }}=\frac{3}{2}=m$;
When $x \rightarrow+\infty$, $y \rightarrow 2=M$ (i.e., $\lim _{x \rightarrow+\infty} y=2$). Hence, the answer is $(A)$.