illusiia

2021-08-16

$\underset{x\to 1}{lim}\left(\frac{{x}^{4}-4}{{x}^{2}-3x+2}\right)$=..

wheezym

If $\underset{x\to a-}{lim}f\left(x\right)\ne \underset{x\to a+}{lim}f\left(x\right)$then the limit does not exist
$⇒\underset{x\to 1+}{lim}\left(\frac{{x}^{4}-4}{{x}^{2}-3x+2}\right)=\underset{x\to 1+}{lim}\left(\left({x}^{4}-4\right)\cdot {\left({x}^{2}-3x+2\right)}^{-1}\right)=\underset{x\to 1+}{lim}\left(\left({x}^{4}-4\right)\cdot \underset{x\to 1+}{lim}{\left({x}^{2}-3x+2\right)}^{-1}\right)=\left(-3\right)\left(-\mathrm{\infty }\right)=\mathrm{\infty }$
$⇒\underset{x\to 1-}{lim}\left(\frac{{x}^{4}-4}{{x}^{2}-3x+2}\right)=\underset{x\to 1+}{lim}\left(\left({x}^{4}-4\right)\cdot \underset{x\to 1+}{lim}{\left({x}^{2}-3x+2\right)}^{-1}\right)=\underset{x\to 1-}{lim}\left({x}^{4}-4\right)\cdot \underset{x\to 1-}{lim}\left({\left({x}^{2}-3x+2\right)}^{-1}\right)=\left(-3\right)\cdot \mathrm{\infty }=-\mathrm{\infty }$
Therefore, $\underset{x\to 1}{lim}\left(\frac{{x}^{4}-4}{{x}^{2}-3x+2}\right)=÷ere\ge s$

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