Question

$\underset{x\to 0}{\lim }\frac{a^x+a^{-x}-2}{x^2}$

Answer 1

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