1. ## Limit with negative radical

Hello,

I'm trying to figure out a problem with a negative radical here. I know there are a few different ways of doing it, but I can't figure it out yet.

$
\lim_{x\to 4} = \sqrt{(x^2-16)}
$

2. $\lim_{x\to 4} \sqrt{x^2-16} = \sqrt{4^2-16} = \sqrt{0} = 0$

3. So...Direct substitution works here?

4. Works for me. Although I don't think the function cannot be
differentiated at x = 4

5. $\sqrt{x^2- 4}$ is continuous for all $|x|\ge 2$. And the definition of "continuous at x= a" is [tex]\lim_{x\rightarrow a} f(x)= f(a). Yes, "direct substitution" works.

In fact, all functions that can be written as "simple formulas" are continuous where ever they are defined.

("Almost all" functions are never continuous but continuous functions are so nice our ways of writing formulas lead to continous functions.)