Calculate limit with modulus

Question

Calculate $$\lim_{x \to x^-_o} \frac{x^2+x-2}{|x+2|}$$when $x_o = -2$

how can I work with the absolute value in this situation?

$$\lim_{x \to -2^-} \frac{(x+2)(x-1)}{|x+2|}$$

answer of limit= $-\frac{3}{2}$ (it was wrong in the book)

Answer 1

$x\to -2^-$ means that $x $ come closer to $-2$ by the left side, or

$x $ stays less than $-2$

thus $x <-2 \iff x+2 <0$

in this case $$|x+2|=-(x+2)$$

After simplification by the common factor $(x+2)$, the function becomes

$$\frac {x-1}{-1}=1-x $$

the limit is $$1-(-2)=3$$

Answer 2

Recall that $$|x+2| = \begin{cases} -(x+2), & \text{if } x<2 \\ x+2, & \text{if } x\ge2 \end{cases}$$ Since it has a $-2^-$, you know which case you have to do.

You should also NOT be getting the answer which you provided. $-\dfrac 32$ is incorrect.

Answer 3

Note that since $x+2<0$

$$\lim_{x \to -2^-} \frac{(x+2)(x-1)}{|x+2|}=\lim_{x \to -2^-} \frac{(x+2)(x-1)}{-(x+2)}=\lim_{x \to -2^-} -x+1=3$$