How can $\lim\limits_{\theta\to0} \theta^{\frac1x -1} \tan(\theta^{\frac1x})$ be evaluated? $$
\lim_{\theta\to0} \theta^{\frac1x -1} \tan(\theta^{\frac1x}) \;\;\;\;\; (x > 1)
$$
I've tried L'Hôpital's rule with $\theta$ in the denominator, but successive applications seems to only lead to more complex expressions. Interestingly, it seems that each application of L'Hôpital's rule will produce another limit to which L'Hôpital's is applicable. 
Can I use this fact somehow to analytically evaluate the limit?
 A: It is not necessary, but I find things easier to see if I let $w=\theta^{1/x}$. Then $\theta=w^x$, and we end up looking at $w^{1-x}\tan w$. 
We want to find
$$\lim_{w\to 0^+} w^{1-x}\tan w.$$
Rewrite this as
$$\lim_{w\to 0^+} w^{2-x} \frac{\tan w}{w}.$$
Now things will likely be clear. The answer is $x$-dependent.  
Remark: We could instead write immediately that our function is
$$\theta^{\frac{2}{x}-1} \frac{\tan(\theta^{\frac{1}{x}})}{\theta^{\frac{1}{x}}},$$ and proceed immediately to the conclusion. However, I think that the preliminary "cleaning up" of the answer helps one see things more clearly. 
A: Let $y=\theta^{1/x}$ ($y^x=\theta$), so as $\theta\to0$, then $y\to0$. Now
Note that
$$
\theta^{1/x-1}=\theta^{1/x}\theta^{-1}=yy^{-x}=y^{1-x}
$$
Thus the limit becomes
$$
\lim_{y\to0}y^{1-x}\tan y
$$
A: The first thing to realize is this limit doesn't exist. That's because $\theta \to 0$ allows negative values of $\theta.$ For such $\theta,$ $\theta^{1/x}$ is not defined for most values of $x.$ So we need to change that limit sign to $\lim_{\theta \to 0^+}.$
That out of the way, note the expression equals
$$\tag 1 \theta^{(2/x)-1}\frac{\tan \theta^{1/x}}{\theta^{1/x}}.$$
As $\theta \to 0^+, \theta^{1/x}\to 0.$ Recalling $\lim_{u\to 0} (\tan u)/u = 1,$ (this follows from $(\sin u)/u \to 1$), we see the fraction in $(1)$  $\to 1$ as $\theta\to 0^+.$ So we don't have to worry about it.
We're left thinking about $\theta^{(2/x)-1}.$ This blows up if $(2/x)-1< 0,$ is identically $1$ if $(2/x)-1 = 0,$ and $\to 0$ if $(2/x)-1>0.$ Thus the limit in $(1)$ doesn't exist if $x>2,$ equals $1$ if $x=2,$ and equals $0$ if $1<x<2.$ 
