# 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?

• The function is $$f(x)=\frac{x^2-36}{\sqrt{x^2-12x+36}},$$ which is not defined for $x=6$. When you simplify you have to take that into account, apart from $\sqrt{a^2}=|a|.$ – Vladimir Vargas Nov 9 '14 at 17:56
• right, but what i dont get is how to get to the two different left hand and right hand limits (they're apparently -12 and 12, i just can't see why!) – the boy that could Nov 9 '14 at 18:05

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 $$\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}=\lim_{x\to6^+}\dfrac{(x+6)(x-6)}{(x-6)}=\lim_{x\to6^+}\phantom{+}(x+6)$$

• right, so then i would get ((x+6)(x-6))/(|x-6|), how exactly do i calculate the limit of that? thanks again for the quick response – the boy that could Nov 9 '14 at 18:03
• your explanation is great, but i don't get why we can just ignore the positive sign for the left hand limit, and ignore the negative sign when calculating the right hand limit. in my mind, i keep making two separate cases for both lhl and the rhl. – the boy that could Nov 9 '14 at 18:09
• Sorry, I don't understand your question. We did not ignore any sign, can you explain what you mean? – Jimmy R. Nov 9 '14 at 18:11
• like, why don't we have two different cases when calculating for the left hand limit (one for the negative sign in front of the denominator, and another for the positive sign), and then do the same for the calculation of the right hand side limit..? this is all after you remove the square root – the boy that could Nov 9 '14 at 18:15
• Oh, ok. Because for the LHL you have that x<6 (since x approaches 6 from below) so you have only the negative sign. Then for the RHL x approaches 6 from above, so you have the case x>6 with the positive sign. Is it clear now? – Jimmy R. Nov 9 '14 at 18:16

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 $|x-6|$.

• even though i gave stefanos the answer, your post really helped me. i'd vote it up, but i dont have the rep! – the boy that could Nov 9 '14 at 18:23
• @theboythatcould That's not a problem, the important thing is that it helped you! :-) – Hakim Nov 9 '14 at 18:25