What is $\lim _{ x\rightarrow 0 } \frac { f(x) -x }{ x^2 }$? Given that
$$f(x)=8x-f(3x)-\sin^2(2x),$$ 
find 
$$\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 }$$ 

$$f(x)=8x-f(3x)-\sin^2(2x)\Longleftrightarrow f(3x) + f(x) = 8x-\sin^2(2x)\\\Longleftrightarrow (f(3x)-3x) + (f(x) -x) = 4x-\sin^2(2x) $$ 
$$\Longleftrightarrow \frac{f(3x)-3x}{9x^2} + \frac{f(x) -x}{9x^2} = \frac{4x-\sin^2(2x)}{9x^2}$$
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. 
