<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] \begin{align*}
x&\ge C \\
f(x) = \dfrac{2x}{x + 1} &= 2\left(1 - \dfrac{1}{x + 1}\right)\\
x&\in [C,\infty)\\
x+1&\in[C+1, \infty)\\
\frac{1}{x+1}&\in\Bigg(0,\frac{1}{C+1}\Bigg]\\
1-\frac{1}{x+1}&\in \Bigg[1-\frac{1}{C+1}, 1\Bigg)\\
\dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right)&\in\Bigg[\frac{2C}{C+1}, 2\Bigg)\\
\end{align*}
So you have $$\frac{2C}{C+1}\le f(x) < 2$$
Alternative:
Since you have already (and correctly) found the lower bound we'll focus on the upper bound. We know that the function is monotonic, since 
$$f'(x) = \frac{2}{(x+1)^2} > 0$$
Hence, we know that the function attains it's maximum (if any) as $x\to \infty$.
Hence we find $$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{2x}{x+1} = \lim_{x\to\infty} \frac{2}{1+\frac{1}{x}} = \fbox{2}$$
Hence, $f(x)<2$.
</s>