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I found this limit within the Calculus Single Variable book from Thomas. $ \lim _{x \to -2^-} (x+3) \frac{|x+2|}{(x+2)}$

This is how I'm trying: First of all, we need to found where the absolute value will apply. $|x+2|$ $x+2=0$ $x=-2$ $ -3+2<0 $ $ -1+2>0 $

So the function will change of sign in this interval: $(-\infty, -2)$ Then I´m trying to solve by substitution but im stuck with the |x+2|, I don't know what to do within the absolute value :(. Thanks.

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    Divide one by the other, at $x=-2.0004$. There, top is equal to ?. Bottom is equal to ?.2011-11-21

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The definition of absolute value is: $|a| = \left\{\begin{array}{ll} a & \text{if }a\geq 0,\\ -a & \text{if }a\lt 0. \end{array}\right.$

That means that (using $x+2$ for $a$): $|x+2| = \left\{\begin{array}{ll} x+2 &\text{if }x+2\geq 0,\\ -(x+2) & \text{if }x+2\lt 0. \end{array}\right.$

When is $x+2\geq 0$? When $x\geq -2$. When is $x+2\lt 0$? When $x\lt -2$. So we can rewrite the above as: $|x+2| = \left\{\begin{array}{ll} x+2 & \text{if }x\geq -2,\\ -(x+2) & \text{if }x\lt -2. \end{array}\right.$

When you take the limit as $x\to -2^{-}$, you are considering values of $x$ that are very close to and less than $-2$. So for those values of $x$, you will have $|x+2| = -(x+2)$.

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    If you are evaluating the limit, you never actually plug in $x=-2$: you only consider what happens as $x$ gets close to, but is not equal to, $-2$. If $x$ is not equal to $-2$, the denominator is not "indeterminate". Since $x$ is not *actually* equal to $-2$, then $\frac{|x+2|}{x+2} = \frac{-(x+2)}{x+2} = -1\quad\text{if }x\lt -2.$2011-11-21