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Question Number 192999 by Mastermind last updated on 01/Jun/23

$$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{following}\mathrm{functions}$$$$\mathrm{are}\mathrm{continous}\mathrm{on}\mathrm{a}\mathrm{close}\mathrm{interval}$$$$[0,1].$$$$\mathrm{f}\left(\mathrm{x}\right)=\{{}_{3\mathrm{x}=1}^{\frac{{\mathrm{x}}^2+\mathrm{x}-2}{\mathrm{x}-1}\mathrm{x}\ne 1}$$$$$$$$\mathrm{Help}!$$

Answered by MM42 last updated on 02/Jun/23

$${lim}_{x\to 0^{+}}f\left(x\right)=f\left(0\right)=2$$$${lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1^{-}}\frac{(x-1)(x+2)}{x-1}=3\mathrm{\&}f\left(1\right)=3$$$$\mathrm{\forall }c\in (0,1);{lim}_{x\to c}f\left(x\right)=f\left(c\right)$$$$therfore;fiscontinouson[0,1]$$

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