Question

$\underset{x\to 0}{\lim }\frac{8^x-4^x-2^x+1}{x^2}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{8^x-4^x-2^x+1}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\left(2^x\right)}^3-{\left(2^x\right)}^2-2^x+1}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\left(2^x\right)}^2\left(2^x-1\right)-1\left(2^x-1\right)}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left(2^{2x}-1\right)\left(2^x-1\right)}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\left(2^{2x}-1\right)}{2x}\times \left(\frac{2^x-1}{x}\right)\right] \\ =2\log 2\times \log 2 \\ =\log {\left(2\right)}^2\times \log 2 \\ =\left(\log 4\right)\times \left(\log 2\right) \end{gathered}$$