# $\lim_{x\to6^-}\big[\frac{\sqrt{36-x^2}}{x-6}\big]$ using L'Hopital's rule?

Disclaimer: this is a homework problem, I at least need a bit of direction, I can't seem to get anywhere, I must be overlooking something.

$$\lim_{x\to6^-}\biggl[\frac{\sqrt{36-x^2}}{x-6}\biggr]$$ I've tried applying L'Hopital's rule since we end up with an indeterminate form if we evaluate at $$x=6$$

I took the first, second, and third derivative but the denominator will always force division by 0. I'm not sure what to do. I'd love to put in the first and second derivatives into this question but, I'm a bit sloppy with the 'Ol Mathjax!

EDIT: Found the big wet mistake: For L'Hopital's rule, I have been taking the derivative of the ENTIRE function, instead of the derivative of the numerator, divided by the derivative of the denominator! Don't do that folks!

• Restricted to that fancy R, Idk how you got that lol. All real numbers, no complex – Cotton Headed Ninnymuggins Oct 31 at 6:08
• This has nothing to do with ceiling-and-floor-functions. – Brian Drake Nov 17 at 10:28
• For future reference, it is better to include your own attempt in the question, especially for homework problems. Had you done so, the other users here would have quickly spotted your mistake and could have pointed it out, instead of doing your homework for you. – Brian Drake Nov 17 at 10:34
• “Restricted to that fancy R, Idk how you got that lol. All real numbers, no complex” I assume you mean that $x$ is a real number ($\mathbb{R}$), that is, $x \in \mathbb{R}$. Note that the real numbers are a subset of the complex numbers ($\mathbb{C}$), so “no complex” is technically incorrect. Example markup: $x \in \mathbb{R}$ – Brian Drake Nov 17 at 10:45

I think you can do it without L'Hospital Rule. Try to multiply with $$\frac{\sqrt{36-x^2}}{\sqrt{36-x^2}}$$. You will then get $$\frac{36-x^2}{(x-6)(\sqrt{36-x^2})}$$. Then you can decompose $$36-x^2$$ to $$(6-x)(6+x)$$ and eliminate the $$x-6$$. You can then apply the limit to get $$-\infty$$.

• Isn’t that still division by 0, I just don’t see the proof here, idk what I’m not seeing – Cotton Headed Ninnymuggins Oct 31 at 6:10
• Yes, but now it is an integer divided by , whose limit will go to infinity (or in this case since the integer is negative, negative infinity). L'Hospital Rule is only applied for equations in the form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$. – Michael Teguh Laksana Oct 31 at 6:12
• I appreciate the help, I really do – Cotton Headed Ninnymuggins Oct 31 at 6:13
• @CottonHeadedNinnymuggins Michael Teguh Laksana’s answer is good, but their explanation is not good. The key point is that the new limit is a nonzero number (doesn’t matter whether it’s an integer) divided by $0$ and therefore $\pm \infty$. To choose the correct sign, we must look at the signs of both numerator and denominator for $x \neq 6$ (the details depend on exactly how you eliminate the common factor). Alternatively, look at the original expression in the question, which is clearly negative. – Brian Drake Nov 17 at 10:03

The domain gives $$-6\leq x<6,$$ which says that our function is not defined for $$x\rightarrow6$$ and we can't say about the limit.

We can try to calculate the limit for $$x\rightarrow6^-$$.

By the L'Hopital's rule we obtain: $$\lim_{x\rightarrow6^-}\frac{\sqrt{36-x^2}}{x-6}=\lim_{x\rightarrow6^-}\frac{\frac{-x}{\sqrt{36-x^2}}}{1}=-\infty.$$

Solution without L'Hopital:

$$\lim_{x\rightarrow6^-}\frac{\sqrt{36-x^2}}{x-6}=-\lim_{x\rightarrow6^-}\sqrt{\frac{6+x}{6-x}}=-\infty.$$

• +1 for actually answering the question, which the accepted answer does not do! – Brian Drake Nov 17 at 10:31

The answer is -infinity. It's not in 0/0form neither in infinity/infinite form.

To get that form divide both numerator and denominator by sqrt(6-x) and u will get the desirable answer

• It is in $0/0$ form. Dividing the numerator and denominator by $\sqrt{6 - x}$ gives a numerator of $\sqrt{6 + x}$, so after this division, it is not in $0/0$ form. – Brian Drake Nov 17 at 9:42