# What is the limit of the Box-Cox transformation as lambda goes to 0? [duplicate]

I am running into this problem and was told this could help.

• The Box-Cox transformation with parameter $\lambda$ is $$x\to \int_1^x y^{\lambda-1}\mathrm{d}y.$$ There's no need to take a limit: simply set $\lambda=0.$ – whuber Dec 17 '19 at 23:52

For parameter values $$\lambda \neq 0$$ the Box-Cox transformation is:

$$y^{(\lambda)} = \frac{y^\lambda-1}{\lambda}.$$

Taking limits and applying L'Hôpital's rule gives:

\begin{aligned} \lim_{\lambda \rightarrow 0} y^{(\lambda)} &= \lim_{\lambda \rightarrow 0} \frac{y^\lambda-1}{\lambda} \\[6pt] &= \lim_{\lambda \rightarrow 0} \frac{(\ln y) y^\lambda}{1} \\[6pt] &= \ln y \times \lim_{\lambda \rightarrow 0} y^\lambda \\[6pt] &= \ln y. \\[6pt] \end{aligned}

In order to preserve continuity of the function, by convention, we define the Box-Cox transform at $$\lambda = 0$$ by its limit:

$$y^{(0)} \equiv \lim_{\lambda \rightarrow 0} y^{(\lambda)} = \ln y.$$

This is one possible way to look at the Box-Cox transformation. As whuber points out in the comment above, the Box-Cox transformation can actually be defined for all cases by the following integral:

$$y^{(\lambda)} \equiv \int \limits_1^y r^{\lambda-1} dr.$$

This definition is well-defined for all $$\lambda \in \mathbb{R}$$ and reduces directly to the stipulated form when $$\lambda =0$$.