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.
1 Answer
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 $|x+2| = -(x+2)$.
