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Question Number 117646 by bemath last updated on 13/Oct/20

Commented by bemath last updated on 13/Oct/20

$$\mathrm{without}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$

Answered by bobhans last updated on 13/Oct/20

$$\underset{x\to 1}{\lim }\left[\frac{1-2\mathrm{x}+\sqrt{\mathrm{x}}}{1-\sqrt{2\mathrm{x}-{\mathrm{x}}^2}}\right]\times \frac{1+\sqrt{2\mathrm{x}-{\mathrm{x}}^2}}{1+\sqrt{2\mathrm{x}-{\mathrm{x}}^2}}=$$$$2\times \underset{x\to 1}{\lim }\frac{(1+\sqrt{\mathrm{x}})-2\mathrm{x}}{{\mathrm{x}}^2-2\mathrm{x}+1}=2\times \underset{x\to 1}{\lim }\frac{(1+\sqrt{\mathrm{x}})-2\mathrm{x}}{{(\mathrm{x}-1)}^2}$$$$2\times \underset{x\to 1}{\lim }\frac{(1+\sqrt{\mathrm{x}})-2\mathrm{x}}{{(\sqrt{\mathrm{x}}+1)}^2{(\sqrt{\mathrm{x}}-1)}^2}=\frac{1}{2}\times \underset{x\to 1}{\lim }\frac{(1+\sqrt{\mathrm{x}})-2\mathrm{x}}{{(\sqrt{\mathrm{x}}-1)}^2}$$$$-\frac{1}{2}\times \underset{x\to 1}{\lim }\frac{2{\left(\sqrt{\mathrm{x}}\right)}^2-\sqrt{\mathrm{x}}-1}{{(\sqrt{\mathrm{x}}-1)}^2}=$$$$-\frac{1}{2}\times \underset{x\to 1}{\lim }\frac{(2\sqrt{\mathrm{x}}+1)(\sqrt{\mathrm{x}}-1)}{{(\sqrt{\mathrm{x}}-1)}^2}=$$$$-\frac{1}{2}\underset{x\to 1}{\lim }\frac{2\sqrt{\mathrm{x}}+1}{\sqrt{\mathrm{x}}-1}=\pm \mathrm{\infty }\mathrm{or}\mathrm{doesnot}\mathrm{exist}$$

Answered by MJS_new last updated on 13/Oct/20

$$0⩽x<1\Rightarrow 1-2x+\sqrt{x}>0\wedge 1-\sqrt{2x-x^2}>0$$$$1<x⩽2\Rightarrow 1-2x+\sqrt{x}<0\wedge 1-\sqrt{2x-x^2}>0$$$$\Rightarrow$$$$\underset{x\to 1^{-}}{\lim }\frac{1-2x+\sqrt{x}}{1-\sqrt{2x-x^2}}>0$$$$\underset{x\to 1^{+}}{\lim }\frac{1-2x+\sqrt{x}}{1-\sqrt{2x-x^2}}<0$$$$\Rightarrow \mathrm{the}\mathrm{limit}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

Commented by bemath last updated on 13/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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