= Solution
For the real-valued expression, first consider the <one-sided limit> $x\to0^+$. Its <natural logarithm> is $x\ln(\ln(1+x))$. Since $\ln(1+x)/x\to1$,
$$
x\ln(\ln(1+x))=x\ln x+x\ln\left(\frac{\ln(1+x)}x\right)\longrightarrow0,
$$
using $x\ln x\to0$. The continuity of the <exponential function> therefore gives \b[right-hand limit $1$].
The printed two-sided notation needs qualification: $\ln(1+x)<0$ for $-1<x<0$, so raising it to an arbitrary real power does not define a real-valued function throughout a left neighbourhood of zero. There is consequently no ordinary two-sided real limit on an open punctured neighbourhood with the expression as written. The value $1$ is the natural right-hand interpretation.
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