Does this epsilon proof work? $x>K ⟹|()−|<$
$f(x)=\dfrac{1}{x}+\log \left(\dfrac{x}{1+x}\right)\rightarrow 0$ with $x \rightarrow \infty$.
Using this inequality:
$\frac{x-1}{x}\leq \log(x) \leq x-1$
$\dfrac{\dfrac{x}{1+x}-1}{\dfrac{x}{1+x}} \leq \log \left(\frac{x}{1+x}\right) \leq \frac{x}{1+x}-1$
$-\frac{1}{x} \leq \log \left(\frac{x}{1+x}\right) \leq \frac{-1}{1+x}$
$0 \leq \log \left(\frac{x}{1+x}\right) +\frac{1}{x} \leq \frac{-1}{1+x}+\frac{1}{x}$
$\log \left(\frac{x}{1+x}\right) +\frac{1}{x} \leq \frac{1}{(1+x)x}$
set $\epsilon=\dfrac{1}{(1+K)K}$
$\epsilon=\dfrac{1}{(1+K)K} \Leftrightarrow K=\frac{-\epsilon+\sqrt{\epsilon^2+4\epsilon}}{2\epsilon}$
The calculation is all fine, but it falls short at the end. Setting $\varepsilon$ to be a function of $K$ is a misstep; you don't get to determine what form $\varepsilon$ takes, as it is an arbitrary positive number.
The next step, where you give $K$ as a function of $\varepsilon$ is better, however, there is a small logical hole where you actually prove that $$x > \frac{-\varepsilon+\sqrt{\varepsilon^2+4\varepsilon}}{2\varepsilon} \implies |f(x) - 0| < \varepsilon.$$ All your argument shows is that when $x = K$, then $|f(x) - L| \le \frac{1}{x(x + 1)} = \varepsilon$, which is not the same thing. How do you know that larger values of $x$ won't result in larger values of $|f(x) - L|$?
Essentially, to fix this approach, you need to show that $\frac{1}{x(x + 1)}$ is strictly monotone decreasing (which it is, as it is the reciprocal of a positive strictly monotone increasing function). That way, you know that $$x \ge K \implies |f(x) - L| \le \frac{1}{x(x + 1)} < \frac{1}{K(K + 1)} = \varepsilon.$$
Alternatively, you could use the squeeze theorem, or compare with an easier to work with upper bound, e.g. $\frac{1}{x^2}$ instead of $\frac{1}{x(x + 1)}$.