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Question Number 144413 by imjagoll last updated on 25/Jun/21

$$\underset{x\to 1}{\lim }[\frac{1}{4-4\sqrt{\mathrm{x}}}-\frac{1}{5-5\sqrt[5]{\mathrm{x}}}]=?$$

Answered by Olaf_Thorendsen last updated on 25/Jun/21

$$f\left(x\right)=\frac{1}{4-4\sqrt{x}}-\frac{1}{5-5\sqrt[5]{x}}$$$$\mathrm{Let}x=X^{10}$$$$f\left(X^{10}\right)=\frac{1}{4-4\sqrt{X^{10}}}-\frac{1}{5-5\sqrt[5]{X^{10}}}$$$$f\left(X^{10}\right)=\frac{1}{4-4X^5}-\frac{1}{5-5X^2}$$$$f\left(X^{10}\right)=\frac{1}{4(1-X)(1+X+X^2+X^3+X^4)}-\frac{1}{5(1-X)(1+X)}$$$$f\left(X^{10}\right)=\frac{5(1+X)-4(1+X+X^2+X^3+X^4)}{20(1-X)(1+X)(1+X+X^2+X^3+X^4)}$$$$f\left(X^{10}\right)=\frac{1+X-4X^2(1+X+X^2)}{20(1-X)(1+X)(1+X+X^2+X^3+X^4)}$$$$\underset{x\to 1}{\lim }f\left(x\right)=\underset{X\to 1}{\lim }f\left(X^{10}\right)=-\frac{10}{200\times 0^{+/-}}$$$$\underset{x\to 1}{\lim }f\left(x\right)=\pm \mathrm{\infty }$$

Answered by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{4-4\sqrt{\mathrm{x}}}-\frac{1}{5-5\left({}^5\sqrt{\mathrm{x}}\right)}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{4(1-\sqrt{\mathrm{x}})}-\frac{1}{5(1-{\mathrm{x}}^{\frac{1}{5}})}$$$$\mathrm{changement}\mathrm{x}-1=\mathrm{t}\mathrm{give}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right)=\frac{1}{4(1-\sqrt{1+\mathrm{t}})}-\frac{1}{5(1-{(1+\mathrm{t})}^{\frac{1}{5}})}$$$$(\mathrm{t}\to 0)\Rightarrow \mathrm{g}\left(\mathrm{t}\right)\sim \frac{1}{4(1-(1+\frac{\mathrm{t}}{2}))}-\frac{1}{5(1-(1+\frac{\mathrm{t}}{5}))}$$$$\Rightarrow \mathrm{g}\left(\mathrm{t}\right)\sim \frac{1}{-2\mathrm{t}}-\frac{1}{-\mathrm{t}}=\frac{1}{\mathrm{t}}-\frac{1}{2\mathrm{t}}=\frac{1}{2\mathrm{t}}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=\mathrm{\infty }={\lim }_{\mathrm{x}\to 1}\mathrm{g}\left(\mathrm{x}\right)$$

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