# Find the limits. Write infty or -infty where appropriate. lim_{xrightarrow0^-}frac{x^2-3x+2}{x^3-4x}

Ava-May Nelson 2021-01-31 Answered
Find the limits. Write $\mathrm{\infty }$ or $-\mathrm{\infty }$ where appropriate. $\underset{x\to {0}^{-}}{lim}\frac{{x}^{2}-3x+2}{{x}^{3}-4x}$
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## Expert Answer

Bella
Answered 2021-02-01 Author has 81 answers

First let's write the limit
$\underset{x\to {0}^{-}}{lim}\frac{{x}^{2}-3x+2}{{x}^{3}-4x}$
$=\underset{h\to 0}{lim}\frac{{x}^{2}-3x+2}{{x}^{3}-4x}$
$=\underset{h\to 0}{lim}\frac{-{h}^{2}-3\left(-h\right)+2}{-{h}^{3}+4h}$
$=\underset{h\to 0}{lim}\frac{{h}^{2}+3g+2}{-{h}^{3}+4h}$
Then, since h tends to 0, it must be fractional and less than 1
As $h<1$
${h}^{3}$
Hence, ${h}^{3}<4h$
$4h-{h}^{3}>0$
Hence, the denominator in the limit above approaches ${0}^{+}$
$Limit=\frac{0+0+2}{{0}^{+}}=\mathrm{\infty }$
Result:$\mathrm{\infty }$

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