calculus - Evaluating $lim_{x to 0^+} left [ frac pi 2 + tan x - arctan frac 1x right]^{frac 1 {ln x}}$ without de L'Hopital?

Making use of the identity:
$$\arctan x + \arctan \frac 1x = \left\{\begin{align}
\pi/2, x \gt 0\\
-\pi/2, x \lt 0
\end{align}\right.$$

I evaluated the limit as follows:
$$\begin{align}
\lim_{x \to 0^+} \left [ \frac \pi 2 + \tan x - \arctan \frac 1x \right]^{\frac 1 {\ln x}} &= \lim_{x \to 0^+} \left ( \tan x + \arctan x \right)^{\frac 1 {\ln x}} =\\

&= e^{\ \displaystyle\lim_{x \to 0^+} \frac{\ln(2x + \mathcal{o}(x))}{\ln x}} = \tag{$\star$} \\
&= e^1 = e
\end{align}$$

At $(\star)$ I Taylor-expandend $\tan x = x + \mathcal{o}(x)$ and $\arctan x = x + \mathcal{o}(x)$, then I had to use de L'Hopital to finish. I also tried not expanding but at the end I had to use de L'Hopital anyway...

I really dislike using de L'Hopital (I find it too mechanical), but here I don't see any alternatives. So, what are the alternatives to evaluate:
$$\lim_{x \to 0^+}\frac{\ln(\tan x + \arctan x)}{\ln x}$$
or the general case:
$$\lim_{x \to 0^+}\frac{\ln \alpha x}{\ln x},\ \text{with}\ \alpha \in \mathbb{R}^+$$

without de L'Hopital?

Or, if you have other ideas to evaluate the original limit, I'll read them with interest!

Answer

$$\lim_{x\to0^+}\frac{\ln(kx)}{\ln x}=\lim_{x\to0^+}1+\frac{\ln(k)}{\ln x}=1$$
Since as ${x\to0^+}$, $\ln x\to-\infty$