anonymous
  • anonymous
...
Mathematics
schrodinger
  • schrodinger
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apoorvk
  • apoorvk
....
apoorvk
  • apoorvk
. . . . .
lgbasallote
  • lgbasallote
???

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anonymous
  • anonymous
calculate this limit \[\lim_{x \rightarrow 2^{+}}\sqrt{x-2}\]
apoorvk
  • apoorvk
aah so you have a question, sorry, will solve it then.
anonymous
  • anonymous
zero from your eyes replace x by 2
anonymous
  • anonymous
then see if this one exists \[\lim_{x \rightarrow 2^{-}}\sqrt{x-2}\]
anonymous
  • anonymous
no because the domain of \(\sqrt{x-2}\) is \(x\geq 2\) so you cannot replace x by values less than 2
apoorvk
  • 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'.
apoorvk
  • apoorvk
LHL of 2+ is 2. RHL is a wee bit more that 2.000...001, you can say 2.000....002
apoorvk
  • 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
anonymous
  • anonymous
yup. thank u. i think the person tha wrote the answers to this exercise list was on crack.
apoorvk
  • apoorvk
Oh sorry I didn't see the other question you had posted, I was only explaining the earlier problem. wait.
apoorvk
  • apoorvk
In the second problem, Limit doesnt exist, exactly for the same reasons that they do in the first problem
apoorvk
  • apoorvk
@nickymarden
anonymous
  • anonymous
yeah, thats the answer i got, but here it says the limit is 1. This professor is mental lol
anonymous
  • anonymous
doesn't know how to type
apoorvk
  • apoorvk
nope, limit can't be possibly one. there may have been a printing or typing error, not necessarily your prof's fault.
TuringTest
  • TuringTest
this is like the fifth problem I've seen of yours with the incorrect answer I feel for you :P
anonymous
  • anonymous
seee, everytime i think i'm going crazy..but thank you guys, AGAIN.
TuringTest
  • TuringTest
very welcome!

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