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Question Number 162138 by LEKOUMA last updated on 27/Dec/21

Answered by Ar Brandon last updated on 27/Dec/21

$$\mathcal{L}=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{1-\ln (e-x)}$$$$=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{1-\ln e-\ln (1-\frac{x}{e})}$$$$=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{\frac{x}{e}}=\underset{x\to 0^{+}}{\lim }\frac{e}{\sqrt{x}}\to +\mathrm{\infty }$$

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