Kaycee Roche

2021-11-06

Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ∞ or −∞, and describe the corresponding vertical asymptote.
$\underset{x\to -{2}^{±}}{lim}\frac{4{x}^{2}+7}{{x}^{3}+8}$

bahaistag

Given function is:
$f\left(x\right)=\frac{4{x}^{2}+7}{{x}^{3}+8}$
$\underset{x\to -{2}^{+}}{lim}\left(\frac{4{x}^{2}+7}{{x}^{3}+8}\right)=\frac{\underset{x\to -{2}^{+}}{lim}\left(4{x}^{2}+7\right)}{\underset{x\to -{2}^{+}}{lim}\left({x}^{3}+8\right)}$
Put, x=-2
$=\frac{4{\left(-2\right)}^{2}+7}{{\left(-2\right)}^{3}+8}$
$=\frac{4×4+7}{-8+8}$
$=\frac{23}{0}$
$=\mathrm{\infty }$
and
$\underset{x\to -{2}^{-}}{lim}\left(\frac{4{x}^{2}+7}{{x}^{3}+8}\right)=\frac{\underset{x\to -{2}^{-}}{lim}\left(4{x}^{2}+7\right)}{\underset{x\to -{2}^{-}}{lim}\left({x}^{3}+8\right)}$
Put, x=-2
$=\frac{4{\left(-2\right)}^{2}+7}{{\left(-2\right)}^{3}+8}$
$=\frac{4×4+7}{-8+8}$
$=\frac{23}{0}$
$=-\mathrm{\infty }$
$\therefore x=-2$ is vertical asymptote.

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