Q:

Proof that Rényi divergence = KL divergence when $\alpha \rightarrow 1$

Kullback–Leibler divergence between two parametrized distributions is defined as:
$$
D_{KL}(q(\theta) || p(\theta)) = \int q(\theta) \log \frac{q(\theta)}{p(\theta)} d\theta
$$
Rényi divergence is defined as:
$$
D_{\alpha}(q(\theta) || p(\theta)) = \frac{1}{\alpha-1} \log\int p(\theta)^\alpha q(\theta)^{1-\alpha} d\theta
$$
It is known that the KL divergence is a particular case of Rényi divergence when $\alpha \rightarrow 1$. 
But what is the proof for that?

A:

Take the limit with L'Hopital:
\begin{align*}
\lim_{\alpha\rightarrow 1} D_\alpha(p||q) &=
\lim_{\alpha\rightarrow 1} \frac{1}{\alpha-1}\log\int p(\theta)^\alpha q(\theta)^{1-\alpha}\,d\theta \\
&= 
\lim_{\alpha\rightarrow 1} \frac{\partial}{\partial \alpha} \log\int p(\theta)^\alpha q(\theta)^{1-\alpha}\,d\theta \\[2mm]
&= \lim_{\alpha\rightarrow 1} \frac{\displaystyle\int p(\theta)^\alpha q(\theta)^{1-\alpha} [\log(p(\theta))-\log(q(\theta))] \,d\theta}{\displaystyle \int p(\theta)^\alpha q(\theta)^{1-\alpha}\,d\theta}  \\
&=    \frac{\displaystyle \int p(\theta)\log(p(\theta)/q(\theta))\,d\theta}{\displaystyle \int p(\theta)\,d\theta} \\[2mm]
&=\int p(\theta)\log\left(\frac{p(\theta)}{q(\theta)}\right)\,d\theta \\[2mm]
&= D_{\text{KL}}(p||q)
\end{align*}

