<s>[INST] Is it possible find the function $f(n)$? Is it possible find the function $f(n)$ ?
 $$\lim_{n\to\infty}\frac{2^{8 n - 3} - 2^{2n-1}- 3}{2^{4 f(n) - 1} - 2^{f(n)} -1} = 9$$
 [/INST] $\lim \limits_{n \to \infty} \frac{2^{8n-3}}{2^{4f(n)-1)}} =9 $ since both are leading terms in numerator and denominator,
So $2^{8n-3-4f(n)+1} = 9$ or $2^{8n-3-4f(n)+1} = 2^{\log_2(9)}$ which means that 
$8n-2-4f(n) = \log_2(9) $ so $4f(n) = 8n-2-\log_2(9)$ so $f(n) = 2n-\frac{1}{2} -\frac{1}{4} \log_2(9) \approx 2n -0.5788$.
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