Question

${lim}_{x\to 0}{\left\{\frac{e^x+e^{-x}-2}{x^2}\right\}}^{\frac{1}{x^2}}$

Answer 1

${lim}_{x\to 0}{\left\{\frac{e^x+e^{-x}-2}{x^2}\right\}}^{\frac{1}{x^2}}={lim}_{x\to 0}{\{1+\frac{e^x+e^{-x}-2}{x^2}-1\}}^{\left(\frac{1}{x^2}\right)}=e^{{lim}_{x\to 0}}(\frac{e^x+e^{-x}-2}{x^2}-1)\times \left(\frac{1}{x^2}\right)\because e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \propto e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots \propto e^x+e^{-x}=2+\frac{2x^2}{2!}+\frac{2x^4}{4!}+\cdots \propto =e^{{lim}_{x\to 0}}(\frac{2+\frac{2x^2}{2!}+\frac{2x^4}{4!}\cdots \propto -2}{x^2}-1)\times \left(\frac{1}{x^2}\right)=e^{{lim}_{x\to 0}}(\frac{\frac{2x^2}{2!}+\frac{2x^4}{4!}+\cdots \propto }{x^4}-\frac{1}{x^2})=e^{{lim}_{x\to 0}}\left(\frac{x^2+\frac{x^4}{12}+\cdots \propto -x^2}{x^4}\right)=e^{\frac{1}{12}}$