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???

calculate this limit \[\lim_{x \rightarrow 2^{+}}\sqrt{x-2}\]

aah so you have a question, sorry, will solve it then.

zero from your eyes
replace x by 2

then see if this one exists \[\lim_{x \rightarrow 2^{-}}\sqrt{x-2}\]

no because the domain of \(\sqrt{x-2}\) is \(x\geq 2\) so you cannot replace x by values less than 2

LHL of 2+ is 2. RHL is a wee bit more that 2.000...001, you can say 2.000....002

yup. thank u. i think the person tha wrote the answers to this exercise list was on crack.

yeah, thats the answer i got, but here it says the limit is 1. This professor is mental lol

doesn't know how to type

this is like the fifth problem I've seen of yours with the incorrect answer
I feel for you :P

seee, everytime i think i'm going crazy..but thank you guys, AGAIN.

very welcome!