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## Answered question

2021-02-22

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

### Answer & Explanation

odgovoreh

Skilled2021-02-23Added 107 answers

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
$\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\underset{x\to 1+}{lim}\frac{\left({x}^{4}-4\right)}{\left({x}^{2}-3x+2\right)}=\underset{x\to 1+}{lim}\left(\left({x}^{4}-4\right)\left({x}^{2}-3x+2{\right)}^{1}\right)$
$=\underset{x\to 1+}{lim}\left({x}^{4}-4\right)×\underset{x\to 1+}{lim}\left(\left({x}^{2}-3x+2{\right)}^{-1}\right)$
$=\left(-3\right)\left(-\mathrm{\infty }\right)$
$=\mathrm{\infty }$
$\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\underset{x\to 1-}{lim}\frac{\left({x}^{4}-4\right)}{\left({x}^{2}-3x+2\right)}=\underset{x\to 1-}{lim}\left(\left({x}^{4}-4\right)\left({x}^{2}-3x+2{\right)}^{1}\right)$ $=\underset{x\to 1-}{lim}\left({x}^{4}-4\right)×\underset{x\to 1-}{lim}\left(\left({x}^{2}-3x+2{\right)}^{-1}\right)$ Conclution: the $lim$ is diverges

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