Let $X\_n$ be a random variable distributed on $A\_n:=\{1, \ldots, n\}$ and $g\_n\colon A\_n \to A\_n$ such that $\Pr\big(X\_n \neq g\_n(X\_n)\big) \to 0$. Putting $Y\_n=g(X\_n)$ then by Fano's inequality $$\frac{H(X\_n\mid Y\_n)}{\log n} \to 0,$$ which can be written $$\frac{H(X\_n\mid Y\_n)}{H(X\_n)} \to 0 \qquad (\ast)$$ in the particular case when $X\_n$ is uniformly distributed on $A\_n$. 


I'm looking for a case for which $H(X\_n)\to \infty$ and $(\ast)$ fails.



order of convergence of the conditional entropy
Let $n=k+k^k$ and let $X$ take each element $1\le j\le k$ with probability $1/k-1/k^2$. Let the remaining $k^k$ elements have probability $1/k^{k+1}$. 
Let $g(i)=\min(k+1,i)$. 


We now have $\mathbb P(X\ne Y)=\mathbb P(X>k)=k^k/k^{k+1}=1/k$, which converges to 0 as required.


We have 
$$
\begin{split}
H(X)&=-k(1/k-1/k^2)\log (1/k-1/k^2)-k^k/k^{k+1}\log (1/k^{k+1}) \cr
&= (1-1/k)(\log k-\log(1-1/k))+(1/k)(k+1)\log k\cr
&=2\log k+O(\log k/k),
\end{split}
$$
which converges to $\infty$ as required. 


Finally, given $Y$, $X$ is known if $Y$ is in the range $\{1,\ldots,k\}$. But if $Y$ takes the value $k+1$ (which occurs with probability $1/k$), then $X$ takes one of $k^k$ values with equal probability. Hence $H(X|Y)=(1/k)\log k^k=\log k$. 


In particular, we have
$$\frac{H(X|Y)}{H(X)}\to\frac 12.
$$


