Let $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.$.

If for some $\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21$, then $\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\frac{|x|^3}{\mathrm{a}}-\left[\frac{x}{\mathrm{a}}\right]\right\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :

$f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.$ <br/><br/>Let $a$ is odd <br/><br/>$$ \begin{aligned} & \Rightarrow f(a)=2 a \\\\ & \Rightarrow f(f(a))=2 a-1 \\\\ & \Rightarrow f(f(f(a)))=2(2 a-1) \end{aligned} $$ <br/><br/>$2(2 a-1)=21$ Not possible for any $a \in N$ <br/><br/>Let $a$ is even <br/><br/>$$ \begin{aligned} & \Rightarrow f(a)=a-1 \\\\ & \Rightarrow f(f(a))=2(a-1) \\\\ & \Rightarrow f(f(f(a)))=2(a-1)-1=2 a-3 \\\\ & 2 a-3=21 \quad \Rightarrow a=12 \end{aligned} $$ <br/><br/>Now <br/><br/>$$ \begin{aligned} & \lim _{x \rightarrow 12^{-}}\left(\frac{|x|^3}{2}-\left[\frac{x}{12}\right]\right) \\\\ & =\lim _{x \rightarrow 12^{-}} \frac{|x|^3}{12}-\lim _{x \rightarrow 12^{-}}\left[\frac{x}{12}\right] \\\\ & =144-0=144 . \end{aligned} $$