LL(k) language and not LL(k) grammar
You can prove by contradiction, assuming there is a value of $k$ that allows you to anticipate the sequence of derivations that is guaranteed to produce a unique tree for any input string, given some unambiguous grammar that accepts the language. 

You then prove that there is a string for which a lookahead  $&gt;k$ will be needed, no matter how the grammar is constructed, because the derivation sequence is affected early in the process. A strategy similar to the pumping lemma for context-free languages comes to mind.

Incidentally, your grammar is equivalent to the following LL(1) grammar:

$\begin{align} S &amp;\to aT \\ 
               T &amp;\to f \mid bXcV \\
               X &amp;\to bXc \mid \epsilon \\
               V &amp;\to d \mid f \end{align}$

Perhaps a more interesting example would be

$\begin{align} S &amp;\to T \mid U \mid V \\ 
               T &amp;\to aTb \mid \epsilon \\
               U &amp;\to aUc \mid \epsilon \\
               V &amp;\to dVc \mid \epsilon \end{align}$

whose language is both not $LL(k)$ and not $RR(k)$.