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Question Number 159332 by 0731619 last updated on 15/Nov/21

Commented by bobhans last updated on 16/Nov/21

$$\left(1\right){\lim }_{\mathrm{x}\to 1^{-}}\sqrt{\mathrm{x}-1}=\mathrm{undefined}$$$${\lim }_{\mathrm{x}\to 1^{+}}\sqrt{\mathrm{x}-1}=0$$$$\left(2\right){\lim }_{\mathrm{x}\to 1^{-}}\sqrt{1-\mathrm{x}}=0$$$${\lim }_{\mathrm{x}\to 1^{+}}\sqrt{1-\mathrm{x}}=\mathrm{undefined}$$

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