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Question Number 106883 by john santu last updated on 07/Aug/20

$$◊\mathrm{JS}⧫$$$$\underset{x\to -1}{\lim }\frac{\sqrt{{\mathrm{x}}^2-1}+1-\sqrt{-\mathrm{x}}}{\sqrt{\mathrm{x}+1}}?$$

Answered by bemath last updated on 07/Aug/20

$$@\mathrm{bemath}@$$$$\underset{x\to -1}{\lim }\frac{\sqrt{\mathrm{x}+1}\sqrt{\mathrm{x}-1}+1-\sqrt{-\mathrm{x}}}{\sqrt{\mathrm{x}+1}}=$$$$\underset{x\to -1}{\lim }\sqrt{\mathrm{x}-1}+\frac{1-\sqrt{-\mathrm{x}}}{\sqrt{\mathrm{x}+1}}\times \frac{1+\sqrt{-\mathrm{x}}}{1+\sqrt{-\mathrm{x}}}=$$$$\underset{x\to -1}{\lim }\sqrt{\mathrm{x}-1}+\frac{1+\mathrm{x}}{\sqrt{\mathrm{x}+1}(1+\sqrt{-\mathrm{x}})}=$$$$\underset{x\to -1}{\lim }(\sqrt{\mathrm{x}-1}+\frac{\sqrt{\mathrm{x}+1}}{1+\sqrt{-\mathrm{x}}})=\mathrm{i}\sqrt{2}?$$$$$$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

$$\underset{x\to -1}{\lim }\sqrt{x-1}+\frac{1-\sqrt{-x}}{\sqrt{x+1}}$$$$\underset{x\to -1}{\lim }i\sqrt{2}+\frac{1+x}{\sqrt{x+1}}.\frac{1}{1+\sqrt{x}}=\underset{x\to -1}{\lim }i\sqrt{2}+\frac{\sqrt{x+1}}{1+\sqrt{x}}=i\sqrt{2}$$

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