anonymous 4 years ago ...

1. apoorvk

....

2. apoorvk

. . . . .

3. anonymous

???

4. anonymous

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

5. apoorvk

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

6. anonymous

zero from your eyes replace x by 2

7. anonymous

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

8. anonymous

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

9. apoorvk

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'.

10. apoorvk

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

11. apoorvk

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

12. anonymous

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

13. apoorvk

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

14. apoorvk

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

15. apoorvk

@nickymarden

16. anonymous

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

17. anonymous

doesn't know how to type

18. apoorvk

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

19. TuringTest

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

20. anonymous

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

21. TuringTest

very welcome!