# 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!

• will you see taylor? – RE60K Sep 16 '14 at 16:53
• @Aditya: Can you elaborate, please? I'm afraid I don't understand. – rubik Sep 16 '14 at 16:57
• will taylor expansion suffice "without L'Hospital"? – RE60K Sep 16 '14 at 17:01
• @Aditya: Sure, I've used it myself. – rubik Sep 16 '14 at 17:57

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