<s>[INST] Limit of square root function at $x \to 6$ 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?
 [/INST] 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 $$\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}=\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{(x-6)}=\lim_{x\to6^+}\phantom{+}(x+6)$$
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