The Rényi entropy for a probability density function $f$ with dominating measure $\mu$ of order $\alpha>0$ is defined as $$H_\alpha(f)={1 \over {\alpha-1}}\log\int f^\alpha d\mu.$$ If $f$ is smooth enough, as $\alpha \to 1$, the limit is the classical Shannon entropy $$H_1(f)=\int f\log f d\mu.$$ The Rényi entropy naturally defines a class of divergence $$M_\alpha(f,g)={1 \over {\alpha-1}}\log\int f^\alpha g^{1-\alpha} d\mu.$$ For fixed $f,g$ smooth enough, the limit divergence as we push $\alpha \to 1$ is the classical Kullback-Leibler divergence $$M_1(f,g)=\int f\log {f \over g} d\mu.$$ Now let's consider a slightly more complicated scenario: If $||g_\alpha/g||_\infty=1+o(1)$, then \begin{equation*} \begin{split} M_\alpha(f,g_\alpha)&={1 \over {\alpha-1}}\log\int f^\alpha g^{1-\alpha}\bigg(\frac{g_\alpha}{g}\bigg)^{1-\alpha} d\mu\\ &={1 \over {\alpha-1}}\bigg(\log \int f^\alpha g^{1-\alpha} d\mu+(1-\alpha)\log(1+o(1)\bigg)\\ &\to M_1(f,g). \end{split} \end{equation*} However, at the entropy level, the simple argument wouldn't apply to show $$H_\alpha(f_\alpha) \to H_1(f)$$ even if $||f_\alpha/f||_\infty=1+o(1)$. This is also tricky when think about divergence in its first entry. Is this something special about the asymmetry of the divergence $M_\alpha(\cdot,\cdot)$? Is there counterexample for which the convergence fails if $||f_\alpha/f||_\infty=1+o(1)$? Or any possible convergence conditions on $f_\alpha \to f$, without controlling the rate of convergence in terms of $\alpha$ that guarantees the convergence?