Originally Posted by

**Guy** I think the question is just asking for you to show that, for fixed y, the function $\displaystyle y^{(\lambda)}$ is continuous in $\displaystyle \lambda$. For $\displaystyle \lambda \ne 0 $ this trivial. To show continuity at $\displaystyle \lambda = 0$ it suffices to show that

$\displaystyle \displaystyle

\lim_{\lambda \to 0} y^{(\lambda)} = \log y

$

which is easy. The simpliest way of going about this is to define $\displaystyle f(\lambda) = y^\lambda$ and write $\displaystyle y^{(\lambda)} = \frac{f(\lambda) - f(0)}{\lambda - 0}$ which is a form that we all recognize.