Let a straight line ("line 1") in the $xy$-plane have one end fixed at the origin $(0,0)$, and the other at a variable point $(x, x – \log(x))$ on the curve $y = x – \log(x)$. The domain of $x$ is the positive Real Numbers.
Let $A$ be the angle between line 1 and the positive $x$-axis, then $\tan(A) = (x – \log(x))/x$ which approaches 1 from below as $x \rightarrow \infty$. As $A = \pi/4$ is the only solution of $\tan(A) = 1$ in the positive quadrant, it seems line 1 rotates towards the straight line $y = x$ as $x \rightarrow \infty$.
I do not see any problem so far (although there may be one). My problem comes when I try to picture what is happening.
The gap between the curve $y = x – \log(x)$ and the straight line $y = x$, is $\log(x)$ which increases as $x$ increases. It therefore seems line 1 should be rotating away from the straight line $y = x$ instead of towards it.
These conclusions are contradictory. I would appreciate any help in pointing out where the error is.
This is not contradictory. This means that your intuition (the second argument) is wrong.
Let's denote $f(x)=x-\ln(x)$, $O$ the origin, $B(x,x)$ a point on the line $y=x$, $C(x,f(x))$ a point on the curve $y=f(x)$, and $P(x,0)$ on the $x$-axis.
Your question is: why the angle $\theta(x)=\angle BOC$ isn't diverging to $-\infty$ when the distance $BC$ tends to $\infty$?
That happens because the distance $OP$ is increasing way faster than the distance $BC$?
A clearer situation is when $f(x)=\frac{1}{2}x$. Clearly, the distance $BC$ tends to $\infty$, but the angle $\theta(x)$ is constant.