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Question Number 168014 by mathlove last updated on 31/Mar/22

$$\underset{x\to 0}{\lim }\frac{5e^x+5e^{-x}-10}{4x^2}=?$$

Answered by som(math1967) last updated on 01/Apr/22

$$\frac{5e^x+\frac{5}{e^x}-10}{4x^2}$$$$=\frac{5e^x-5-\frac{5}{e^x}(e^x-1)}{4x^2}$$$$=\frac{5(e^x-1)-\frac{5}{e^x}(e^x-1)}{4x^2}$$$$=\frac{(e^x-1)(5-\frac{5}{e^x})}{4x^2}$$$$=\frac{5(e^x-1)(e^x-1)}{4x^2{e}^x}$$$$\therefore \underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{5}{4}\times {\left(\frac{e^x-1}{x}\right)}^2\times \frac{1}{e^x}$$$$=\frac{5}{4}\times 1^2\times 1=\frac{5}{4}$$

Commented by som(math1967) last updated on 01/Apr/22

$$\mathit{o}\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{w}\mathit{a}\mathit{y}$$$$\frac{5}{4}\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{e^x+e^{-x}-2}{x^2}$$$$=\frac{5}{4}\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{{\mathit{e}}^{2\mathit{x}}+1-2{\mathit{e}}^{\mathit{x}}}{{\mathit{e}}^{\mathit{x}}\times {\mathit{x}}^2}$$$$=\frac{5}{4}\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}{\left(\frac{{\mathit{e}}^{\mathit{x}}-1}{\mathit{x}}\right)}^2\times \frac{1}{{\mathit{e}}^{\mathit{x}}}$$$$=\frac{5}{4}\times 1^2\times 1=\frac{5}{4}$$$$$$

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