I need to find the left and right limit at x=1(if they exist)
$\displaystyle sqrt1-x^2$
I'm not sure where to begin with the sqaure root.
Thanks
Jason
$\displaystyle lim_{x\rightarrow 1^-} \sqrt{1-x^2}=0$ I think this is clear
$\displaystyle lim_{x\rightarrow 1^+} \sqrt{1-x^2}=dose...not.....exist$
this dose not exist as you know square root of negative numbers undefined right when I take the limit of the function when x approach 1 from right it is like I take a value bigger than 1 just a littile like this 1.0000000000000000000000000000000001 so the value under the square root become negative and this not allow