Question

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Answer 1

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Answer 2

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Answer 3

???

Answer 4

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

Answer 5

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

Answer 6

zero from your eyes
replace x by 2

Answer 7

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

Answer 8

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

Answer 9

well, for 2+, limit exists. put just replace the x=2+ (since we do not have to overcome any indeterminant form)

now 2+ is basically a no. a very wee bit larger than 2. that is about... "2.000000...001"

so square root of (0.0000000...0001)
is almost negligible, that is zero.


So, '0'.

Answer 10

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

Answer 11

so when 2- isnt coming into the picture, so we don't have a prob there with domain or what lies under the square root.


You're welcome

Answer 12

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

Answer 13

Oh sorry I didn't see the other question you had posted, I was only explaining the earlier problem. wait.

Answer 14

In the second problem, Limit doesnt exist, exactly for the same reasons that they do in the first problem

Answer 15

@nickymarden

Answer 16

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

Answer 17

doesn't know how to type

Answer 18

nope, limit can't be possibly one. there may have been a printing or typing error, not necessarily your prof's fault.

Answer 19

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

Answer 20

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

Answer 21

very welcome!