<s>[INST] What is $\lim _{ x\rightarrow 0 } \frac { f(x) -x }{ x^2 }$? Given that
$$f(x)=8x-f(3x)-\sin^2(2x),$$ 
find 
$$\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 }$$ 
 [/INST] Note that
$$ f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3). $$
Then from $f(x)=8x-f(3x)-\sin^2(2x)$, one has
\begin{eqnarray}
&&\bigg[f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3)\bigg]+\bigg[f(0)+3f'(0)x+\frac{9}2f''(0)x^2+O(x^3)\bigg]\\
&=&8x-\sin^2(2x)
\end{eqnarray}
which gives
$$ 2f(0)+4f'(0)x+5f''(0)x^2+O(x^3)=8x-\sin^2(2x). $$
Noting that $\sin(2x)\approx 2x$ and hence
$$f(0)=0,f'(0)=2,f''(0)=-\frac{4}{5}. $$
So
$$ \lim_{x\to0}\frac{f(x)-x}{x^2}=\frac{f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3)-x}{x^2}=\lim_{x\to0}\frac{x-\frac{2}{5}x^2}{x^2}=DNE.$$
</s>