Question

The limit of √x-4 as x approaches 4

Answer 1

careful here

Answer 2

Ok

Answer 3

note that while \(\sqrt{4-4}=\sqrt{0}=0\) the limit means the limit from both directions, right and left
so while \(\lim_{x\to 4^+}\sqrt{x-4}=0\) is good, you cannot approach \(4\) from numbers below \(4\) becuase if \(x<4\) then \(x-4<0\) and so \(\sqrt{x-4}\) is not a real number

Answer 4

what i am trying to say is that
\[\lim_{x\to 4^-}\sqrt{x-4}\] does not exist, because the expression \(\sqrt{x-4}\) is only defined for \(x\geq 4\)

Answer 5

Ok that makes a lot of sense. Thank you!

Answer 6

yw