Limit question with absolute value: $ \lim_{x\to 4^{\large -}}\large \frac{x-4}{|x-4|} $ How would I solve the following limit, as $\,x\,$ approaches $\,4\,$ from the left?
$$
\lim_{x\to 4^{\large -}}\frac{x-4}{|x-4|}
$$
Do I have  to factor anything?
 A: Hint:  If $x \lt 4, |x-4|=4-x$.  Now you can just divide.
A: As $x \to 4$ from the left,  $\;x \lt 4 \implies x - 4 \lt 0 \implies |x-4|=(4-x) = -(x - 4).\;$  
Substitute $\;-(x - 4)\;$ for $\;|x - 4|\;$ into the original limit, and divide. 
$$\lim_{x\to 4^-}\frac{x-4}{|x-4|} = \; \lim_{x\to 4^-}\frac{(x-4)}{-(x - 4)} = -1.$$
after canceling the common factor in the numerator and denominator.

Similarly, as $x \to 4$ from the right, $x \gt 4 \implies (x - 4) > 0 \implies |x - 4| = (x - 4)$.
So $$\lim_{x\to 4^+}\frac{x-4}{|x-4|} = \; \lim_{x\to 4^+}\frac{(x-4)}{(x - 4)} = 1.$$
A: $$
\lim_{x\to 4^-}\frac{x-4}{|x-4|}=\lim_{x\to 4^-}\frac{x-4}{-(x-4)}=-1
$$
A: Yes, you do have to factor something. But you have to take a little bit of care.
If you want the left-hand limit, you have $x<4$. If $x<4$ then $x-4 < 0$ and so $|x-4|$ is actually the negative of $x-4$. In other words: $|x-4| \equiv -(x-4)$ for all $x<4$. Hence:
$$\frac{x-4}{|x-4|} \equiv \frac{x-4}{-(x-4)} \equiv -1$$
for all $x<4$. It follows that the limit is also $-1$. A similar argument shows that $|x-4|\equiv x-4$ for all $x>4$ and so the right-hand limit is $+1$.
