What is $\lim _{ x\rightarrow 0 } \frac { f(x) -x }{ x^2 }$? Given that
$$f(x)=8x-f(3x)-\sin^2(2x),$$ 
find 
$$\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 }$$ 
 A: $$f(x)=8x-f(3x)-\sin^2(2x)\Longleftrightarrow f(3x) + f(x) = 8x-\sin^2(2x)\\\Longleftrightarrow (f(3x)-3x) + (f(x) -x) = 4x-\sin^2(2x) $$ 
$$\Longleftrightarrow \frac{f(3x)-3x}{9x^2} + \frac{f(x) -x}{9x^2} = \frac{4x-\sin^2(2x)}{9x^2}$$
Let $$\ell = \lim_{x\to 0}\frac{f(x) -x}{x^2}$$
Then, 
$$\lim_{x\to 0}\frac{4x-\sin^2(2x)}{9x^2}= \lim_{x\to 0}\frac{f(3x)-3x}{9x^2} +\frac{f(x) -x}{9x^2}=\lim_{x\to 0}\frac{f(3x)-3x}{(3x)^2} +\frac19\frac{f(x) -x}{x^2}= \frac{10}{9}\ell $$
Since $$\lim_{x\to 0}\frac{f(3x)-3x}{9x^2}=\ell$$
Hence, 
$$\lim_{x\to 0}\frac{f(x) -x}{x^2}=\ell =\frac{1}{10} \lim_{x\to 0}\frac{4x-\sin^2(2x)}{x^2} \sim \frac{1}{10} \lim_{x\to 0}\frac{4x-4x^2}{x^2}  $$
But this latest limit does not exists. 
A: Suppose $\lim_{x\to 0}(f(x)-x)/x^2 = L \in \mathbb R.$ Then $f(x) = x + O(x^2).$ This implies $f(3x) = 3x + O(x^2).$ Putting this into the given equation and simplifying then gives
$$4x = O(x^2),$$
a contradiction. Thus such an $L$ does not exist.
A: 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
$$ \lim_{x\to0}\frac{f(x)-x}{x^2}=\frac{f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3)-x}{x^2}=\lim_{x\to0}\frac{x-\frac{2}{5}x^2}{x^2}=DNE.$$
