For a fixed delta > 0, evaluate the limit as rho tends to infinity of (2 arctan(rho/delta)/pi)^rho, and decide whether ln of that quantity minus 1 can ever be defined for real rho.

Step-by-step solution

1. 1

   Pin down the base for finite $\rho$. For every finite $\rho>0$ and fixed $\delta>0$ the argument $\rho/\delta$ is finite, so $\arctan\frac{\rho}{\delta}<\frac{\pi}{2}$ strictly — the value $\frac{\pi}{2}$ is only a horizontal asymptote, never attained. Therefore

   $0<\frac{2\arctan\frac{\rho}{\delta}}{\pi}<1$

   This is the crux: the base approaches $1$ from below, so raising it to a large power pulls it down rather than leaving it at $1$.

2. 2

   Rewrite the base as $1$ minus a small quantity. Use the identity $\arctan u+\arctan\frac{1}{u}=\frac{\pi}{2}$ for $u>0$:

   $\arctan\frac{\rho}{\delta}=\frac{\pi}{2}-\arctan\frac{\delta}{\rho}\qquad\Longrightarrow\qquad \frac{2\arctan\frac{\rho}{\delta}}{\pi}=1-\frac{2}{\pi}\arctan\frac{\delta}{\rho}$

   As $\rho\to+\infty$ we have $\frac{\delta}{\rho}\to 0$ and $\arctan u\sim u$, so the base behaves like $1-\frac{2\delta}{\pi\rho}$.

3. 3

   Apply the classical exponential limit. Since $\lim_{\rho\to\infty}\left(1-\frac{c}{\rho}\right)^{\rho}=e^{-c}$, taking $c=\frac{2\delta}{\pi}$ gives

   $\lim_{\rho\to+\infty}\left(\frac{2\arctan\frac{\rho}{\delta}}{\pi}\right)^{\rho}=e^{-2\delta/\pi}$

   The familiar warning $\left(1-\frac1n\right)^{n}\to e^{-1}$, not $1$, is exactly the same phenomenon: a base tending to $1$ fights an exponent tending to infinity, and the product $\rho\cdot\frac{2\delta}{\pi\rho}$ decides the winner.

4. 4

   Conclude that the logarithm's argument is negative. Because $\delta>0$ we get $0<e^{-2\delta/\pi}<1$, hence

   $\left(\frac{2\arctan\frac{\rho}{\delta}}{\pi}\right)^{\rho}-1\ \longrightarrow\ e^{-2\delta/\pi}-1<0$

   Moreover, for every finite $\rho>0$ the base is already strictly below $1$, so the bracket is strictly negative — not just in the limit. The real logarithm of a negative number is undefined, so that $\ln$ never makes sense over the reals.

5. 5

   Verify both claims numerically. With $\delta=1$: $e^{-2/\pi}=0.5290778$, while the expression gives $0.5191644$ at $\rho=10$, $0.5280135$ at $\rho=100$, $0.5289707$ at $\rho=10^{3}$ and $0.5290778$ at $\rho=10^{7}$ — converging to the predicted value, clearly not to $1$. With $\delta=3$ the target $e^{-6/\pi}=0.1481012$ is likewise reached. And the bracket evaluates to $-0.2410$, $-0.5000$, $-0.4808$ and $-0.4710$ at $\rho=0.1,1,10,1000$: negative throughout.