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# Evaluate the limit, if it exists. $\displaystyle \lim_{x \to -2}\frac{x + 2}{x^3 + 8}$

## $\frac{1}{12}$

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to evaluate the limits of X plus two over X. Q plus eight. As X approaches negative to be first Do direct substitution. That means we replace X by -2 and we have negative two plus two over negative to be to the third power plus eight. This gives us zero over negative eight plus eight which is also zero. And this value is indeterminate. That means we need to rewrite our function exports to over execute plus eight. Now this we can be right into limit as X approaches negative too. We have X-plus two over. We can factor out execute plus eight into X plus two times X squared minus two. X plus four. And then from here can cancel us the X plus two. And we have limits As X approaches -2 of one over X squared minus two. X plus four. And then from here we Evaluated at x equals -2 and we get one over the square of negative two minus two times negative two plus four. This gives us 1/4 plus four plus four Or we have won over 12. And so this is the value of the limits

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