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  ?
 A: 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 $x=1$.
A: You need to be a bit more careful when manipulating inequalities.
$k_{1} x \leq f(x) \leq k_{2} x$ and $\frac{1}{2}k_{1} x \leq g(x) \leq k_{2} x$. 
Hence we have $-k_{2} x \leq -f(x) \leq -k_{1} x$ and $-k_{2} x \leq -g(x) \leq -\frac{1}{2}k_{1}x $. 
This gives us $f(x)-g(x) \geq k_{1} x-k_{2}x = \left(k_{1}-k_{2}\right)x$.
This also gives us $g(x)-f(x) \geq \frac12 k_{1} x- k_{2}x = \left(\frac12 k_{1}-k_{2}\right)x$.
Hence $|f(x)-g(x)| \geq \max\{\left(k_{1}-k_{2}\right)x, \left(\frac12 k_{1}-k_{2}\right)x\}$.
We have $k_{1}-k_{2}>\frac12 k_{1}-k_{2}$ since $k_1$ is positive.
If $x>0$, this gives $|f(x)-g(x)| \geq \left(k_{1}-k_{2}\right)x$.
However, we have that $k_2\geq k_1$, so it tells us nothing more than $|f(x)-g(x)| \geq 0$. In fact, in can go to zero and then $\frac{f(x)+x}{|f(x)-g(x)|} \to \infty$, provided that the numerator doesn't go to 0. 
