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Question Number 169711 by mnjuly1970 last updated on 06/May/22

$$$$$$provethat:$$$$$$$${lim}_{x\to 0}(\frac{1}{x^2}-\frac{e^x}{{(e^x-1)}^2})=\frac{1}{12}$$$$$$

Commented by infinityaction last updated on 06/May/22

$$ithinkquestioniswrong$$

Commented by cortano1 last updated on 07/May/22

$$no.thequestionright$$

Commented by infinityaction last updated on 07/May/22

$$thankyousir$$$$gotmymistake$$

Answered by qaz last updated on 07/May/22

$$\mathrm{L}=\underset{\mathrm{x}\to 0}{\lim }(\frac{1}{{\mathrm{x}}^2}-\frac{{\mathrm{e}}^{\mathrm{x}}}{{({\mathrm{e}}^{\mathrm{x}}-1)}^2})=\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{2\mathrm{x}}-{2\mathrm{e}}^{\mathrm{x}}+1-{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{x}}^4}$$$${\mathrm{e}}^{2\mathrm{x}}-{2\mathrm{e}}^{\mathrm{x}}+1-{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}$$$$=(1+2\mathrm{x}+\frac{1}{2}{\left(2\mathrm{x}\right)}^2+\frac{1}{6}{\left(2\mathrm{x}\right)}^3+\frac{1}{24}{\left(2\mathrm{x}\right)}^4+...)-2(1+\mathrm{x}+\frac{1}{2}{\mathrm{x}}^2+\frac{1}{6}{\mathrm{x}}^3+\frac{1}{24}{\mathrm{x}}^4+...)+$$$$1-{\mathrm{x}}^2(1+\mathrm{x}+\frac{1}{2}{\mathrm{x}}^2+...)$$$$=(\frac{16}{24}-\frac{2}{24}-\frac{1}{2}){\mathrm{x}}^4+...=\frac{1}{12}{\mathrm{x}}^4+\mathrm{o}\left({\mathrm{x}}^4\right)$$$$\Rightarrow \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{12}{\mathrm{x}}^4+\mathrm{o}\left({\mathrm{x}}^4\right)}{{\mathrm{x}}^4}=\frac{1}{12}$$

Answered by cortano1 last updated on 07/May/22

$$\underset{x\to 0}{\lim }\frac{{(e^x-1)}^2-x^2{e}^x}{x^2{(e^x-1)}^2}$$$$=\underset{x\to 0}{\lim }\frac{{(1+x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^3\right)-1)}^2-x^2(1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+O\left(x^3\right))}{x^2{(1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+O\left(x^3\right)-1)}^2}$$$$=\underset{x\to 0}{\lim }\frac{x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+...-(x^2+x^3+\frac{1}{2}{x}^4+\frac{1}{6}{x}^5+...)}{x^2{(x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+...)}^2}$$$$=\underset{x\to 0}{\lim }\frac{\frac{1}{12}{x}^4+O\left(x^5\right)}{x^4+O\left(x^5\right)}=\frac{1}{12}$$

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