Limit of square root function at $x \to 6$

Question

I'm trying to find the limit of the following function at $x \to 6$:

$$\frac{x^2-36}{\sqrt{x^2-12x+36}}$$

i've simplified it so that it becomes $\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}$, which simplifies to $x+6$.

the problem is that i shouldn't be getting to $x+6$, because then id be able to plug in $6$, and say that the limit exists for the left hand side and the right hand side of $6$, when clearly i can tell from the graph that the limit does not exist.

What am I doing wrong?

Answer 1

Hint: $$\sqrt{(x-6)^2}=|x-6|$$ which is in turn equal to $$|x-6|=\begin{cases}-(x-6)=-x+6, & x<6\\ \phantom{+}(x-6)=\phantom{-}x-6,&x>6\end{cases}$$ This implies that $$\lim_{x\to6^-}\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}=\lim_{x\to6^-}\dfrac{(x+6)(x-6)}{-(x-6)}=\lim_{x\to6^-}-(x+6)$$ but $$\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}=\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{(x-6)}=\lim_{x\to6^+}\phantom{+}(x+6)$$

Answer 2

Hint:
That's because $\sqrt{(x-6)^2}$ doesn't simplify to $x-6$ but to $|x-6|$ which is very different since: $$|x-6|=\begin{cases}x-6, & x\geqslant6\\ -(x-6),&x\leqslant6\end{cases}.$$ So when you will compute the limit from the right and that from the left you'll get different results because of the difference in the sign engendered by $|x-6|$.