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!

  • $\begingroup$ will you see taylor? $\endgroup$ – RE60K Sep 16 '14 at 16:53
  • $\begingroup$ @Aditya: Can you elaborate, please? I'm afraid I don't understand. $\endgroup$ – rubik Sep 16 '14 at 16:57
  • $\begingroup$ will taylor expansion suffice "without L'Hospital"? $\endgroup$ – RE60K Sep 16 '14 at 17:01
  • $\begingroup$ @Aditya: Sure, I've used it myself. $\endgroup$ – 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 ${x\to0^+}$, $\ln x\to-\infty$

  • $\begingroup$ Ah, straightforward, thanks! $\endgroup$ – rubik Sep 16 '14 at 17:54

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