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Question Number 152371 by fotosy2k last updated on 27/Aug/21

Commented by fotosy2k last updated on 27/Aug/21

$$plsbelp$$$$$$

Answered by Olaf_Thorendsen last updated on 27/Aug/21

$$\underset{x\to 2}{\lim }\frac{x-2}{x^2-3x+2}=\underset{x\to 2}{\lim }\frac{x-2}{(x-2)(x-1)}$$$$=\underset{x\to 2}{\lim }\frac{1}{x-1}=1$$

Commented by fotosy2k last updated on 27/Aug/21

$$thankyou$$

Answered by Rasheed.Sindhi last updated on 28/Aug/21

$$\mathbf{AnOther}\mathbf{Approach}$$$$\underset{x\to 2}{\lim }\frac{x-2}{x^2-3x+2}$$$$x=2+h\Rightarrow x\to 2\Rightarrow h\to 0$$$$\underset{h\to 0}{\lim }\frac{2+h-2}{{(2+h)}^2-3(2+h)+2}$$$$\underset{h\to 0}{\lim }\frac{h}{4+4h+h^2-6-3h+2}$$$$\underset{h\to 0}{\lim }\frac{h}{h+h^2}=\underset{h\to h}{\lim }\frac{1}{1+h}=\frac{1}{1+0}=1$$$$$$

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