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Question Number 121074 by bramlexs22 last updated on 05/Nov/20

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

Answered by liberty last updated on 05/Nov/20

$$\underset{x\to 1}{\lim }\frac{\mathrm{x}-1}{\sqrt{(\mathrm{x}-1)(\mathrm{x}+1)}}=\underset{x\to 1}{\lim }\sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}=0$$

Answered by liberty last updated on 05/Nov/20

$${\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 1}{\lim }\frac{\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2-1}}=\underset{x\to 1}{\lim }\frac{1}{\left[\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}\right]}$$$$=\underset{x\to 1}{\lim }\frac{\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}}=0$$

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