Past exam of the mathematical methods II course of the University of Oxford / 2016 / 2 / b / Solution

For the real-valued expression, first consider the one-sided limit $x\to 0^{+}$. Its natural logarithm is $x\ln (\ln (1+x))$. Since $\ln (1+x)/x\to 1$,using $x\ln x\to 0$. The continuity of the exponential function therefore gives 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.