Question: If D4 is acting on the subgroup of its rotations, C4, by conjugation, what are the orbits?

I believe that the orbit of each rotation is itself and its own inverse rotation and nothing else. For the identity, it's just the identity.
Is this correct? What's a good way to show/ conclude this? I determined this through drawing out the actions.

Answer: 
For quickly finding the limit; apply L'Hospital's rule to $$\lim\_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}}$$

To get $$\lim\_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}} =\lim\_{x \to 0} \frac{\frac 1x}{\frac{-1}{2 x \sqrt x}}=\lim\_{x \to 0} \left(-2\sqrt x \right)=0$$
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> **I thought it would be instructive to present a way forward that relies on elementary, pre-calculus tools only. To that end we proceed.**
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To show that $\sqrt x\log(x) \to 0$ as $x\to 0$, we simply exploit the inequality

$$\frac{x-1}{x}\le\log(x)\le x-1 \tag1$$ 

which I showed using only the limit definition of the exponential function and Bernoulli's inequality in [THIS ANSWER](http://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263). Note for $0<x\le 1$, we have from $(1)$

$$\frac{x-1}{x}\le \log(x)\le 0 \tag2$$

Replacing $x$ with $x^\alpha$ in $(2)$, multiplying by $\sqrt x$, and using $\log(x^\alpha)=\alpha \log(x)$, we find that for $0<x\le 1$ and $\alpha >0$

$$\frac{\sqrt x-x^{1/2-\alpha}}\alpha\le \sqrt x\log(x)\le 0$$

Restricting $0<\alpha<1/2$ (e.g., Take $\alpha=1/4$.), applying the squeeze theorem yields coveted limit 

$$\lim{x\to 0}\sqrt x\log(x)=0$$

as was to be shown!

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Note that by enforcing the substitution $x\to x^2$ and using integration by parts, we find that 

$$\begin{align}
\lim\_{\epsilon\to 0}\int\_{\epsilon}^1 \sqrt{x}\log(x)\,dx&=\lim\_{\epsilon\to 0}\left.\left(\frac29 x^{3/2}(3\log(x)-2)\right)\right|\_{\epsilon}^1\\\\
&=\lim\_{\epsilon\to 0}\left(-\frac49 -\frac29 \epsilon^{3/2}(3\log(\epsilon)-2)\right)\\\\
&=-\frac49
\end{align}$$

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> So, the integral converges and the value to which it converges is $-\frac49$.
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