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Question Number 68885 by aliesam last updated on 16/Sep/19

$$if$$$$f\left(x\right)=\frac{\mid x\mid }{x}$$$$g\left(x\right)=x^2-1$$$$$$$$find$$$$$$$$\underset{x\to 1}{lim}f\left(g\left(x\right)\right)$$

Commented by kaivan.ahmadi last updated on 16/Sep/19

$$f\left(g\left(x\right)\right)=f(x^2-1)=\frac{\mid x^2-1\mid }{x^2-1}$$$${lim}_{x\to 1^{+}}\frac{\mid x^2-1\mid }{x^2-1}={lim}_{x\to 1^{+}}\frac{x^2-1}{x^2-1}=1$$$${lim}_{x\to 1^{-}}\frac{\mid x^2-1\mid }{x^2-1}={lim}_{x\to 1^{-}}\frac{-(x^2-1)}{x^2-1}=-1$$$$\Rightarrow {lim}_{x\to 1}f\left(g\left(x\right)\right)isnotexist.$$

Commented by aliesam last updated on 16/Sep/19

$$if$$$$\underset{x\to c}{lim}g\left(x\right)$$$$then$$$$\underset{x\to c}{lim}f\left(g\left(x\right)\right)$$$$=f\left(\underset{x\to c}{lim}g\left(x\right)\right)$$$$=f\left(b\right)$$

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