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

Question

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!

Answer 1

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 $-\infty$.

Answer 2

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

Answer 3

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:

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