<s>[INST] If $x \geq C$, where $C > 0$ is a constant, then what is the least upper bound for $\dfrac{2x}{x + 1}$? The title says it all.
Since
$$f(x) = \dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right),$$
then because $x \geq C$ where $C > 0$, an upper bound is given by
$$\dfrac{2x}{x + 1} < 2.$$
Note that the lower bound is equivalent to
$$x \geq C \iff x + 1 \geq C + 1 \iff \dfrac{1}{x + 1} \leq \dfrac{1}{C + 1}$$ $$\iff 2\left(1 - \dfrac{1}{x + 1}\right) \geq 2\left(1 - \dfrac{1}{C + 1}\right),$$
so that if $2 - \varepsilon$ is an upper bound for $f(x)$ ($\varepsilon > 0$), then $\varepsilon$ satisfies the inequality
$$2 - \varepsilon \geq f(x) \geq 2\left(1 - \dfrac{1}{C + 1}\right) \implies 0 < \varepsilon < \dfrac{2}{C + 1}.$$
My question is essentially whether we can do better than this.
 [/INST] rewrite $f(x)$ in the form $$\frac{2}{1+\frac{1}{x}}$$ and consider the cases $$x>0$$ or $$x<0$$
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