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Question Number 194968 by cortano12 last updated on 21/Jul/23

Answered by tri26112004 last updated on 21/Jul/23

$$=\underset{x\to 0}{\lim }\frac{1}{\sqrt{x+1}+\sqrt{1-x}}.-\underset{x\to 0}{\lim }\frac{\sqrt{x+1}+\sqrt{1-x}-2}{x^2}$$$$=-\frac{1}{2}.\left\{\underset{x\to 0}{\lim }\frac{\frac{1}{2}[\sqrt{4x+4}-(x+2)]+\sqrt{1-x}+\frac{1}{2}x-1}{x^2}\right\}$$$$=-\frac{1}{2}.[\underset{x\to 0}{\lim }\frac{1}{2}.\frac{-1}{\sqrt{4x+4}+x+2}+\underset{x\to 0}{\lim }\frac{1}{4}.\frac{1}{(\frac{1}{2}x-1)-\sqrt{1-x}}]$$$$=-\frac{1}{2}.(\frac{1}{2}.\frac{-1}{4}+\frac{1}{4}.\frac{1}{-2})=\frac{1}{8}$$

Answered by dimentri last updated on 21/Jul/23

$$=\underset{x\to 0}{\lim }\left(\frac{\frac{2x}{\sqrt{x+1}+\sqrt{1-x}}-x}{x^3}\right)$$$$=\frac{1}{2}.\underset{x\to 0}{\lim }\left(\frac{2-\sqrt{1+x}-\sqrt{1-x}}{x^2}\right)$$$$=\frac{1}{2}.\underset{x\to 0}{\lim }\left(\frac{\frac{-x}{1+\sqrt{1+x}}+\frac{x}{1+\sqrt{1-x}}}{x^2}\right)$$$$=\frac{1}{8}.\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\right)$$$$=\frac{1}{8}.\underset{x\to 0}{\lim }\left(\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}\right)$$$$=\frac{1}{8}$$

Commented by mathlove last updated on 22/Jul/23

$$whay$$$$\frac{\sqrt{1+x}-\sqrt{1-x}-x}{x^3}=\frac{\frac{2x}{\sqrt{1+x}+\sqrt{1-x}}-x}{x^3}$$

Commented by dimentri last updated on 22/Jul/23

$$\frac{\sqrt{1+x}-\sqrt{1-x}.\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}-x}{x^3}$$$$=\frac{\frac{(1+x)-(1-x)}{\sqrt{1+x}+\sqrt{1-x}}-x}{x^3}$$

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