Question

lim x-->1(-) sqrt(x-1) = 0
is that right?

Answer 1

no

Answer 2

\[\lim_{x \rightarrow 1^{-}} \sqrt{x-1}\]

Answer 3

why?

Answer 4

if \(x<1\) then \(x-1<0\) and \(\sqrt{x-1}\) is therefore undefined

Answer 5

isn't the domain [-1,infinity) ?
and 1 is in the domain..?

Answer 6

the domain of \(\sqrt{x-1}\) is \([1,\infty)\)

Answer 7

If you're going from the left, it's not defined.

Answer 8

If going from the right, it is.

Answer 9

you will get an imaginary number with this limit

Answer 10

don't forget for \(\sqrt{x}\) you must have \(x\geq 0\) so for \(\sqrt{x-1}\) you need \(x-1\geq 0\) i.e. \(x\geq 1\)

Answer 11

no limit because the function does not exist on the left

Answer 12

\(x\to 1^{-}\) is going from the left

Answer 13

you are correct that 1 is in the domain, but 1 is the left hand endpoint of the domain, not \(-1\)

Answer 14

Going back to the definition of a limit, you need to give them a \(\delta > 0\) and yet anything greater than zero will result in an imaginary \(f(x)\).

Answer 15

the function isn't defined at x = -1.
so basically, there's no function the the left side of x = 1 so there's no limit as x--> 1(-) ?

Answer 16

\[
\forall \epsilon>0 \exists \delta \quad \forall x\quad 0<1-x < \delta \implies |\sqrt{x-1}-0| <\epsilon
\]
I'll be generous and say \(\epsilon = 10^{100}\). Give me any delta that will work.

Answer 17

Make my day =)

Answer 18

-_- sorry i didn't learn that stuff, but i think i get the idea.. :/

Answer 19

if you approach to the 1 by the LEFT, you will get numbers like \[\sqrt{-10-1}=imaginary-->\sqrt{-9-1}=imaginary-->\sqrt{-8-1}=imaginary-->\sqrt{-5-1}=imaginary-->\sqrt{0-1}=imaginary-->\sqrt{0.9999999-1}=imaginary\]

Answer 20

so on without reaching 1

Answer 21

i think i understand it now.
thanks. :)

Answer 22

:D