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

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  • $\begingroup$ 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 $\lambda=0.$ $\endgroup$ – whuber Dec 17 '19 at 23:52
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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:

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

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 $\lambda =0$.

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