Given the curve y = arctan (ln x). 1° Find the angle

Given the curve $y = \arctan\bigl(\ln x\bigr)$.
1° Find the angle at which the curve intersects the $x$-axis.
2° Calculate $\displaystyle\lim_{x\to0^{+}}y'$ and $\displaystyle\lim_{x\to\infty}y'$.

Step 1: The derivative

By the chain rule, with outer function $\arctan u$ and inner function $u = \ln x$: \[y' = \frac{1}{1+\bigl(\ln x\bigr)^{2}}\cdot\frac1x = \frac{1}{x\bigl(1+\ln^{2}x\bigr)}. \tag{1}\]

Note that $y'\gt 0$ throughout $\bigl(0,\infty\bigr)$: the curve rises everywhere.

Step 2: 1° — where the curve meets the $x$-axis

Setting $y = 0$: \[\arctan\bigl(\ln x\bigr) = 0 \;\Longrightarrow\; \ln x = 0 \;\Longrightarrow\; x = 1,\]

so the crossing point is $\bigl(1,0\bigr)$.

Step 3: 1° — the angle there

Substituting $x = 1$ into (1), where $\ln1 = 0$: \[y'\big|_{x=1} = \frac{1}{1\cdot\bigl(1+0\bigr)} = 1. \tag{2}\]

The angle $\varphi$ with the positive $x$-axis satisfies $\tan\varphi = 1$, hence

\[ \varphi = \arctan1 = \frac{\pi}{4} = 45^{\circ}. \tag{3} \]

Step 4: 2° — the slope as $x\to0^{+}$

Here $x\to0^{+}$ pulls the denominator of (1) down while $1+\ln^{2}x\to+\infty$ pushes it up; the substitution $x = e^{-t}$, $t\to+\infty$, settles which effect wins: $y' = \frac{1}{e^{-t}\bigl(1+t^{2}\bigr)} = \frac{e^{t}}{1+t^{2}} \longrightarrow +\infty$, since the exponential outgrows every polynomial (Problem 393). Therefore

\[ \lim_{x\to0^{+}}y' = +\infty. \tag{4} \]

Step 5: 2° — the slope as $x\to+\infty$

Now both factors of the denominator grow, so no competition arises: \[\lim_{x\to+\infty}y' = \lim_{x\to\infty}\frac{1}{x\bigl(1+\ln^{2}x\bigr)} = 0. \tag{5}\]

Step 6: Results

\[1^{\circ}\quad \text{crossing at } \bigl(1,0\bigr) \text{ at the angle } \varphi = 45^{\circ}\]

Step 7: Verification

\[y'(0.01) = 4.5029645; \qquad y'(0.001) = 20.526681 \;\;\checkmark\]

\[y'(100) = 4.5029645\cdot10^{-4}; \qquad y'\!\left(10^{4}\right) = 1.1650888\cdot10^{-6} \;\;\checkmark\]

\[\text{The curve is bounded: } y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\ \text{with both values as horizontal asymptotes} \;\;\checkmark\]