Question

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

Answer 1

${lim}_{x\to 0}\frac{e^x+e^{-x}-2}{x^2}=\frac{0}{0}$ form

Applying L Hospital rule

${lim}_{x\to 0}\frac{e^x+e^{-x}}{2x}=\frac{0}{0}$

Again applying L-Hospital rule

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