Question

$\underset{x\to 0}{\lim }\frac{9^x-2.6^x+4^x}{x^2}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{9^x-2.6^x+4^x}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\left(3^x\right)}^2-2·3^x·2^x+{\left(2^x\right)}^2}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\left(3^x-2^x\right)}^2}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[{\left(\frac{3^x-2^x}{2^x}\right)}^2\times \frac{{\left(2^x\right)}^2}{x^2}\right] \\ =\underset{x\to 0}{\lim }{\left[\frac{{\left({\displaystyle \frac{3}{2}}\right)}^x-1}{x}\right]}^2\times 2^{2x} \\ ={\left[\log \left(\frac{3}{2}\right)\right]}^2\times 2^0 \\ ={\left[\log \left(\frac{3}{2}\right)\right]}^2 \end{gathered}$$