Why is $\lim_{x \to 0} x^{(x^\epsilon)} = 1$ even for $0<\epsilon << 1$ Why is $$\lim_{x \to 0} x^{(x^\epsilon)} = 1$$ even for $$0<\epsilon << 1?$$
The way I see it, in $\lim_{x\to 0} x^x$, the exponent converging to $0$ would seek to make the expression $1$, but the basis conerging to $0$ would seek to make the expression $0$. In the "normal" case, the exponent converging wins. But the exponent converging wins even when the $0<\epsilon <<1$ halts the convergence to $0$ in the exponent dramatically. Is there a good explanation for this behavior?
 A: Since $x^{x^\epsilon}=e^{x^\epsilon \log x}$ and the exponential function is continuous everywhere, we have
$$
{\large{
\lim_{x \rightarrow 0^{+}} x^{x^\epsilon} = 
\lim_{x \rightarrow 0^{+}} e^{x^\epsilon \log x}
=e^{\lim_{x \rightarrow 0^{+}} x^\epsilon \log x}.
}}$$
Now, to evaluate this limit, we can use L'Hopital's rule:
$$
\lim_{x \rightarrow 0^{+}} x^\epsilon \log x
=
\lim_{x \rightarrow 0^{+}} \frac{\log x}{x^{-\epsilon}}
=
\lim_{x \rightarrow 0^{+}} \frac{1/x}{\epsilon x^{-1-\epsilon}}
=
\lim_{x \rightarrow 0^{+}} \frac{1}{\epsilon} x^{\epsilon} = 0.
$$
Since $e^0=1$, we conclude that 
$$
{\large {\lim_{x \rightarrow 0+} x^{x^\epsilon} = 1}}.$$
So, we can see that, as Daniel pointed out in his comment, $x^\epsilon$ goes to zero ``faster" than $\log x$ goes to $-\infty$, regardless of
the value of $\epsilon$, and this results in the limit being $1$.
Added: It is fun to investigate this numerically and/or graphically, because it may require quite small values of $x$ to see this limit happening.  If we differentiate $x^\epsilon \log x$, we can find that it is increasing for $x > e^{-1/\epsilon}$.  If we take, say, $\epsilon = \frac{1}{10}$, we see that $x^\epsilon \log x$ is increasing (i.e., so getting farther from $0$ as we take $x$ smaller and smaller) for $x>0.0000453$ or so.  For smaller $x$, the function is decreasing, and from this point on $x^\epsilon \log x$ gets closer and closer to zero as we make $x$ smaller and smaller.  And this is just with $\epsilon = 1/10$, which is not really all that small.  As a result, if you plot $x^\epsilon \log x$ with graphing software, it may be hard to see this limit happening, unless we zoom in very close to $x=0$, and maybe not even then, as we may run into limits of numerical precision with $\epsilon$ as small as, say, $1/1000$.
A: The fact is that
$$
\lim_{x\to0}x^\varepsilon\log x=0
$$
for every $\varepsilon>0$. Indeed, you can rewrite it as
$$
\lim_{x\to0}x^\varepsilon\log x=
\lim_{x\to0}\frac{1}{\varepsilon}x^\varepsilon\log(x^\varepsilon)
$$
The substitution $t=x^\varepsilon$ reduces it to
$$
\lim_{t\to0}\frac{1}{\varepsilon}t\log t=0
$$
Therefore
$$
\lim_{x\to0}x^{x^\varepsilon}=
\lim_{x\to0}\exp(x^\varepsilon\log x)=\exp0=1
$$
