Proof that $\alpha$-divergence = KL as $\alpha \rightarrow$ 1 Kullback–Leibler divergence between two parametrized distributions is defined as:
$$D_{KL}(p||q) = \int{p(x)\log{\frac{p(x)}{q(x)}}dx}$$
$\alpha$-divergence is defined as:
$$D_{\alpha}(p || q) = \frac{4}{1 - \alpha^2} \left( 1 - \int p(x)^{\frac{1 + \alpha}{2}}q(x)^{\frac{1 - \alpha}{2}}dx\right)$$
how to proof that $D_{KL}(p||q)$ = $D_{\alpha}(p || q)$ when $\alpha \rightarrow$ 1
 A: Let
$$
D_\alpha (p \| q) := \frac{f(\alpha)}{g(\alpha)}
$$
where
$$
f(\alpha) := 1-\int p^{(1+\alpha)/2} q^{(1-\alpha)/2}, \qquad g(\alpha) = \frac{1-\alpha^2}{4},
$$
then note that
$$
\lim_{\alpha \to 1} D_\alpha (p \| q) = \lim_{\alpha \to 1} \frac{f(\alpha)}{g(\alpha)}=\frac{0}{0}
$$
is indeterminate, so we can apply L'Hopital's rule. To this end, we will differentiate both $f$ and $g$ with respect to $\alpha$ and use the fact that
$$
\lim_{\alpha \to 1} D_\alpha (p \| q) = \lim_{\alpha \to 1} \frac{f'(\alpha)}{g'(\alpha)}.
$$
First, it is easy to see that $g'(\alpha) = -\alpha/2$. Then, by the chain rule,
\begin{align*}
f'(\alpha) 
&= \frac{d}{d \alpha} \int p^{(1+\alpha)/2} q^{(1-\alpha)/2}\\
&=  \int \left ( \frac{d}{d \alpha}p^{(1+\alpha)/2} \right) q^{(1-\alpha)/2}
+ \int p^{(1+\alpha)/2} \left ( \frac{d}{d \alpha} q^{(1-\alpha)/2} \right)\\ 
&= \frac{1}{2} \int p^{(1+\alpha)/2}q^{(1-\alpha)/2} \log \frac{p}{q}.
\end{align*}
The last equality follows since
$$
\frac{d}{d \alpha}p^{(1+\alpha)/2} 
= \frac{d}{d \alpha} \exp\left ( \frac{1+\alpha}{2} \log p  \right)
=\frac{1}{2} \exp\left ( \frac{1+\alpha}{2} \log p  \right) \log p  =
\frac{1}{2}  p^{(1+\alpha)/2}  \log p,
$$
and similarly
$$
\frac{d}{d \alpha}q^{(1-\alpha)/2} 
= \frac{d}{d \alpha} \exp\left ( \frac{1-\alpha}{2} \log q  \right)
= -\frac{1}{2}  q^{(1-\alpha)/2}  \log q,
$$
In summary, we have
$$
\lim_{\alpha \to 1} \frac{f'(\alpha)}{g'(\alpha)}
= \lim_{\alpha \to 1} \frac{\frac{1}{2} \int p^{(1+\alpha)/2}q^{(1-\alpha)/2} \log \frac{p}{q}}{-\alpha/2}
= \int p \log \frac{p}{q} = D_{\text{KL}}(p\| q).
$$
Note that a similar argument shows
$$
\lim_{\alpha \to -1} D_\alpha (p \| q) = D_{\text{KL}}(q\| p)
$$
