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Question Number 162139 by LEKOUMA last updated on 27/Dec/21

Answered by alephzero last updated on 27/Dec/21

$$\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{1-\ln (e-x)}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{d}{dx}\left(\sqrt{x}\right)}{\frac{d}{dx}(1-\ln (e-x))}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(\ln (e-x)\right)}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{0-\frac{d}{dx}\left(\ln (e-x)\right)}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{-\left(\frac{d}{dg}\left(\ln g\right)\times \frac{d}{dx}(e-x)\right)}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{-(\frac{1}{g}\times (\frac{d}{dx}\left(e\right)-\frac{d}{dx}\left(x\right))}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{-\left(\frac{1}{g}\times (0-1)\right)}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{-\left(\frac{1}{g}\times (-1)\right)}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{-\left(\frac{1}{e-x}\times (-1)\right)}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{-(-\frac{1}{e-x})}=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{2\sqrt{x}}}{\frac{1}{e-x}}=$$$$=\underset{x\to 0^{+}}{\lim }\frac{e-x}{2\sqrt{x}}=\underset{x\to 0^{+}}{\lim }\left((e-x)\times \frac{1}{2\sqrt{x}}\right)$$$$\underset{x\to 0^{+}}{\lim }(e-x)=e$$$$\underset{x\to 0^{+}}{\lim }\left(\frac{1}{2\sqrt{x}}\right)=+\mathrm{\infty }$$$$e\times (+\mathrm{\infty })=+\mathrm{\infty }$$$$\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{1-\ln (e-x)}=+\mathrm{\infty }$$

Answered by mathmax by abdo last updated on 27/Dec/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}}}{1-\ln (\mathrm{e}-\mathrm{x})}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\ln (\mathrm{e}-\mathrm{x})=\mathrm{t}\Rightarrow$$$$\mathrm{e}-\mathrm{x}={\mathrm{e}}^{\mathrm{t}}\Rightarrow \mathrm{x}=\mathrm{e}-{\mathrm{e}}^{\mathrm{t}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{e}-{\mathrm{e}}^{\mathrm{t}}}}{1-\mathrm{t}}(\mathrm{x}\to 0\Rightarrow \mathrm{t}\to 1)$$$${=}_{1-\mathrm{t}=\mathrm{z}}\frac{\sqrt{\mathrm{e}-{\mathrm{e}}^{1-\mathrm{z}}}}{\mathrm{z}}(\mathrm{z}\to 0)$$$$\mathrm{\Psi }\left(\mathrm{z}\right)=\frac{\sqrt{\mathrm{e}-{\mathrm{e}}^{1-\mathrm{z}}}}{\mathrm{z}}=\frac{\sqrt{\mathrm{e}}\sqrt{1-{\mathrm{e}}^{-\mathrm{z}}}}{\mathrm{z}}=\frac{\sqrt{\mathrm{e}}}{\mathrm{z}}{(1-{\mathrm{e}}^{-\mathrm{z}})}^{\frac{1}{2}}$$$${\mathrm{e}}^{-\mathrm{z}}\sim 1-\mathrm{z}\Rightarrow 1-{\mathrm{e}}^{-\mathrm{z}}\sim \mathrm{z}\Rightarrow \mathrm{\Psi }\left(\mathrm{z}\right)\sim \frac{\sqrt{\mathrm{e}}}{\mathrm{z}}\sqrt{\mathrm{z}}=\frac{\sqrt{\mathrm{e}}}{\sqrt{\mathrm{z}}}\to +\mathrm{\infty }$$$$\Rightarrow {\lim }_{\mathrm{x}\to 0^{+}}\mathrm{f}\left(\mathrm{x}\right)=+\mathrm{\infty }$$

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