Evaluate the limit of (x^2 - x - 6)/|x + 2| as x approaches -2, or show that it does not exist.

Step-by-step solution

1. 1

   Check the form and factor the numerator. Substituting $x=-2$ gives $\frac{4+2-6}{0}=\frac{0}{0}$, an indeterminate form. Factoring the numerator:

   $x^{2}-x-6=(x-3)(x+2),$

   so the expression becomes $\dfrac{(x-3)(x+2)}{|x+2|}$. The factor $(x+2)$ appears above and, in absolute value, below — that is the signal that a one-sided analysis is required.

2. 2

   Understand why the absolute value forbids a single cancellation. For $t\ne0$, $\frac{t}{|t|}$ equals $+1$ when $t>0$ and $-1$ when $t<0$. Here $t=x+2$ changes sign exactly at the limit point, so the quotient jumps between $+1$ and $-1$ no matter how close to $-2$ you look. The two sides must be treated separately.

3. 3

   Compute the left-hand limit. For $x<-2$ we have $x+2<0$, so $|x+2|=-(x+2)$ and

   $\frac{(x-3)(x+2)}{-(x+2)}=-(x-3)=3-x\ \longrightarrow\ 3-(-2)=5.$

   So $\displaystyle\lim_{x\to-2^{-}}=5$.

4. 4

   Compute the right-hand limit. For $x>-2$ we have $x+2>0$, so $|x+2|=x+2$ and

   $\frac{(x-3)(x+2)}{x+2}=x-3\ \longrightarrow\ -2-3=-5.$

   So $\displaystyle\lim_{x\to-2^{+}}=-5$.

5. 5

   Conclude and sanity-check numerically. Since $5\ne-5$, the one-sided limits disagree and the two-sided limit does not exist. Numerically, at $x=-2.001$ the quotient is $5.001$ and at $x=-1.999$ it is $-4.999$ ✓ — the graph has a jump of height $10$ at $x=-2$. Had the denominator been $x+2$ instead of $|x+2|$, the limit would have existed and equalled $-5$.