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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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Hint: $$ |x| = \left\{ \begin{array}{ccc} x & & x \geq 0 \\ -x& & x < 0 \end{array} \right. $$ –  JavaMan Nov 21 '11 at 17:06
    
Hint 2: Split the limit as a product of two limits. You will need JavaMan's hint to evaluate one of them. [The other one is easy.] –  Srivatsan Nov 21 '11 at 17:09
    
Yeah, but do i need to replace the -2 within the absolute value, or replace the x with an arbitrary value given by the interval created before. –  nEAnnam Nov 21 '11 at 17:11
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If you're only interested in the limit from the negative side, can't you just replace $|x+2|$ with $-(x+2)$? –  Mike Nov 21 '11 at 17:27
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Please do not write things like "$-3+2=\mathrm{Negative}$". The number $-1$ is not equal to the word "Negative", and the correct notation "$-3+2<0$" is both shorter and easier. –  Henning Makholm Nov 21 '11 at 17:50

1 Answer 1

up vote 6 down vote accepted

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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Ok understood :), just one thing, what about the indetermination of the denominator? thanks man –  nEAnnam Nov 21 '11 at 17:34
    
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 $$\frac{|x+2|}{x+2} = \frac{-(x+2)}{x+2} = -1\quad\text{if }x\lt -2.$$ –  Arturo Magidin Nov 21 '11 at 17:37
    
solved :) Thank you –  nEAnnam Nov 21 '11 at 17:41

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