Evaluate
lim x tends to 0 ((e^x)+(e^-x) -2)/x^2

Dear Student,

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}-2}{{\mathrm{x}}^2}=\underset{\mathrm{x}\to 0}{\lim }\frac{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}-2\right)}{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left({\mathrm{x}}^2\right)} \left[\frac{0}{0} \mathrm{form}\right] \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{2\mathrm{x}} \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}\right)}}{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}\left(2\mathrm{x}\right)}} \left[\frac{0}{0} \mathrm{form}\right] \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2} \\ =\frac{{\mathrm{e}}^0+{\mathrm{e}}^{-0}}{2} \\ =\frac{1+1}{2} \\ =1 \\ \Rightarrow \underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{-}\mathbf{2}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathbf{1} \end{gathered}$$

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