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Question Number 184903 by aba last updated on 13/Jan/23

$$\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}-2}{{\mathrm{x}}^2}=??$$$$\mathrm{let}\mathrm{x}=2\mathrm{t}\{\begin{array}{l}\mathrm{x}\to 0\\ \mathrm{t}\to 0\end{array}$$$$\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}-2}{{\mathrm{x}}^2}=\underset{\mathrm{t}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}-2}{{4\mathrm{t}}^2}=\frac{1}{4}\underset{\mathrm{t}\to 0}{\lim }{\left(\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{\mathrm{t}}\right)}^2=\frac{1}{4}\underset{\mathrm{t}\to 0}{\lim }{(\frac{{\mathrm{e}}^{\mathrm{t}}-1}{\mathrm{t}}+\frac{{\mathrm{e}}^{-\mathrm{t}}-1}{-\mathrm{t}})}^2=\frac{1}{4}{(1+1)}^2=1$$

Answered by manxsol last updated on 13/Jan/23

$$$$$$1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3{!}_{}}-2=\frac{2\frac{x^2}{2!}+2\frac{x^4}{4!}}{x^2}=1+\frac{x^2}{12}.......$$$${lim}_{x\to 0}1+\frac{x^2}{12}+.....=1$$$$$$$$$$$$$$

Commented by aba last updated on 13/Jan/23

$$\mathrm{nice}$$

Commented by manxsol last updated on 14/Jan/23

$$$$$$\frac{e^{0.0000000001}+e^{-0.0000000001}-2}{{(0.0000000001)}^2}$$$$0.0$$$$resultofservidor$$$$\frac{e^{0.00001}+e^{-0.00001}-2}{0.{00001}^2}$$$$1.0$$$$resultofservidor$$$$$$$$$$$$$$$$$$

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