 # Find the following: \lim_{x\to-2}(\frac{1}{x^2}+\frac{1}{x^3+4}) E Brennan Flores 2021-11-09 Answered
Find the following:
$\underset{x\to -2}{lim}\left(\frac{1}{{x}^{2}}+\frac{1}{{x}^{3}+4}\right)$
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Given,
$\underset{x\to -2}{lim}\left(\frac{1}{{x}^{2}}+\frac{1}{{x}^{3}+4}\right)$
we know $\underset{x\to a}{lim}\left[f\left(x\right)+g\left(x\right)\right]$
$=\underset{x\to a}{lim}f\left(x\right)+\underset{x\to a}{lim}g\left(x\right)$
Therefore
$\underset{x\to -2}{lim}\left(\frac{1}{{x}^{2}}+\frac{1}{{x}^{3}+4}\right)=\underset{x\to -2}{lim}\frac{1}{{x}^{2}}+\underset{x\to -2}{lim}\frac{1}{{x}^{3}+4}$
$=\underset{x\to -2}{lim}\frac{1}{{x}^{2}}+\underset{x\to -2}{lim}\frac{1}{{x}^{3}+4}$
$=\frac{1}{4}+\frac{1}{{\left(-2\right)}^{3}+4}$
$=\frac{1}{4}+\frac{1}{-8+4}$
$=\frac{1}{4}-\frac{1}{4}=0$