# Limit of a function with absolute values

So I've got this limit:

$$\lim_{x\to 3^-} \frac{x^2-9}{|x-3|}$$

My (wrong) answer was zero. I figured that since the numerator approaches zero then regardless of what the denominator was, the whole function would approach zero. However, after looking at the graph I realize that this is not the case, the function approaches $-6$ when $x$ gets closer to $3$ (using values below than 3). How could one solve this without recurring to the graph?

• Note $x<3\Longrightarrow \frac{x^2-9}{|x-3|}=\frac{(x+3)(x-3)}{-(x-3)}$. – Ángel Mario Gallegos May 1 '14 at 22:34

you can factorize the numerator as $(x-3)(x+3)$.
Note that: $|x-3| = 3-x$ when $x$ goes to $3$ from the left hand side and cancel those two, and see what you will get.
• I have a question though, why do you assume right away that $|x-3|=-(x-3)$ when $x$ goes to $3$ from the left? I know it makes sense, I just wanna know how to figure it out quickly. – Juan José Castro May 2 '14 at 0:27
• @Joseph: in general, the values of $|x-a|$ fall into two cases: (1) it equals $x-a$ when $x\geq a$, and (2) it equals $-(x-a)$ when $x\leq a$. That is straight from the definition of absolute value. In this problem, approaching 3 from the left means you're assuming $x<3$, so you're in the second case. – symplectomorphic May 2 '14 at 2:08