<s>[INST] Is this function bounded above? Consider nonconstant functions $f(x), g(x) \neq x$. Suppose there exist positive constants $k_1$ and $k_2$ such that 
$k_{1} x \leq f(x) \leq k_{2} x$ and $\frac{1}{2}k_{1} x \leq g(x) \leq k_{2} x$. Does it follow that $\ \frac{f(x) + x}{\mid f(x) - g(x)\mid}  < \infty$ ?
My attempt: Since $f(x) + x \leq (k_{2} + 1)x$ and $\mid f(x) - g(x)\mid  \geq \frac{1}{2}k_{1}x  $,
                                                                                the upper bound evaluates to $\frac{2k_{2} + 2}{k_{1}} < \infty$, as required  ?
 [/INST] Let
\begin{align}
f(x) &= k x   & \text{where $k_1 < k < k_2, k\neq -1$,} \\
g(x) &= f(x) - \frac12 k_1 x\arctan(x - 1). &
\end{align}
Then
$$
\frac{f(x) +x}{|f(x)-g(x)|} 
= \frac{(k + 1)x}{\frac12 k_1 x \arctan(x - 1)} 
= \frac{2(k+1)}{k_1 \arctan(x - 1)}.
$$
But $\lim_{x\to 1} \arctan(x - 1) = 0$, and $1/\arctan(x - 1)$ 
is unbounded in any neighborhood of $x=1$.
</s>