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Evaluate the limit, if it exists.

$ \displaystyle \lim_{x \to -2}\frac{x + 2}{x^3 + 8} $

$\frac{1}{12}$

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 3

Calculating Limits Using the Limit Laws

Limits

Derivatives

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This is problem number nineteen of the Stuart Calculus eighth edition Section two point three. Usually with a limited it exists, Limited's experts is negative to the quantity X plus two. To melt away the quantity X cubed plus eight, you will attempt to simplify this Lim, and what we're going to to take advantage of is that the denominator is a different is a sum of cubes, and we have a formula for some of cubes. If you have something like a Cuban, let's be cute. This is equivalent to a plus being multiplied by hey, squared kindness, maybe, plus B squared. So what we do is we identify that here we have one X cubed X being the equivalent to A in this case in this example and ate as a cube is the number two cubed. And so we're gonna take to to be the same as being in this case. So we used T formula here for a sum of cubes we have. The first term is X plus be in this case, X plus two. Not to play by a squared or X squared, minus eight times being or x Times two Plans B squared or two squared, which is for we see how we were able to fact already and X Plus two, which cancels with the numerator leaving us with the limit is expressions negative too? Ah, one divided by this quantity X squared by his two x plus four and in order to evaluate is limit. We're going to play a negative, too, here in denominator and we'LL get. But this simplifies to Warren over four plus for plus for which causes a final answer for this limit. The Limited presenting it to of explosives experts to divided by the quantity Thanks Cube was eight is one over two one over twelve.

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