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Question Number 98952 by Ar Brandon last updated on 17/Jun/20

$$\mathrm{Without}\mathrm{using}{\mathrm{L}}^{\prime }\mathrm{H}{\widehat{\mathrm{o}pital}}^{\prime }\mathrm{s}\mathrm{rule}\mathrm{or}{\mathrm{Maclaurin}}^{\prime }\mathrm{s}\mathrm{expansion}$$$$\mathrm{series},\mathrm{find}\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{{\mathrm{xe}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}-1}-1}{\mathrm{x}}$$

Commented by 675480065 last updated on 17/Jun/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{xe}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}-1}-1\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}$$$$\mathrm{from}\mathrm{taylor}\mathrm{series}\mathrm{expansion}$$$${\mathrm{e}}^{\mathrm{x}}=1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+0\left({\mathrm{x}}^2\right)\Rightarrow {\mathrm{xe}}^{\mathrm{x}}=\mathrm{x}+{\mathrm{x}}^2+\frac{{\mathrm{x}}^3}{2}+0\left({\mathrm{x}}^3\right)$$$${\mathrm{e}}^{\mathrm{x}}-1=1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+0\left({\mathrm{x}}^2\right)-1=\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+0\left({\mathrm{x}}^2\right)$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left[\frac{\mathrm{x}(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+0\left({\mathrm{x}}^2\right))}{\mathrm{x}(1+\frac{\mathrm{x}}{2}+0\left(\mathrm{x}\right))}\right]-1$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left[\frac{1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}-1-\frac{\mathrm{x}}{2}+0\left({\mathrm{x}}^2\right)}{(1+\frac{\mathrm{x}}{2}+0\left(\mathrm{x}\right))}\right]=\left[\frac{\mathrm{x}+{\mathrm{x}}^2+0\left({\mathrm{x}}^2\right)}{(2+\mathrm{x}+0\left(\mathrm{x}\right))}\right]$$$$\underset{x\to 0}{\lim }\left(\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}\right)=\underset{x\to 0}{\lim }\left[\frac{1+\mathrm{x}+0\left(\mathrm{x}\right)}{2+\mathrm{x}+0\left(\mathrm{x}\right)}\right]=\frac{1}{2}$$

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