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nickymarden

  • 4 years ago

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  1. apoorvk
    • 4 years ago
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    ....

  2. apoorvk
    • 4 years ago
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    . . . . .

  3. lgbasallote
    • 4 years ago
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    ???

  4. nickymarden
    • 4 years ago
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    calculate this limit \[\lim_{x \rightarrow 2^{+}}\sqrt{x-2}\]

  5. apoorvk
    • 4 years ago
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    aah so you have a question, sorry, will solve it then.

  6. anonymous
    • 4 years ago
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    zero from your eyes replace x by 2

  7. nickymarden
    • 4 years ago
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    then see if this one exists \[\lim_{x \rightarrow 2^{-}}\sqrt{x-2}\]

  8. anonymous
    • 4 years ago
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    no because the domain of \(\sqrt{x-2}\) is \(x\geq 2\) so you cannot replace x by values less than 2

  9. apoorvk
    • 4 years ago
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    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
    • 4 years ago
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    LHL of 2+ is 2. RHL is a wee bit more that 2.000...001, you can say 2.000....002

  11. apoorvk
    • 4 years ago
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    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. nickymarden
    • 4 years ago
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    yup. thank u. i think the person tha wrote the answers to this exercise list was on crack.

  13. apoorvk
    • 4 years ago
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    Oh sorry I didn't see the other question you had posted, I was only explaining the earlier problem. wait.

  14. apoorvk
    • 4 years ago
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    In the second problem, Limit doesnt exist, exactly for the same reasons that they do in the first problem

  15. apoorvk
    • 4 years ago
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    @nickymarden

  16. nickymarden
    • 4 years ago
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    yeah, thats the answer i got, but here it says the limit is 1. This professor is mental lol

  17. nickymarden
    • 4 years ago
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    doesn't know how to type

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

  19. TuringTest
    • 4 years ago
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    this is like the fifth problem I've seen of yours with the incorrect answer I feel for you :P

  20. nickymarden
    • 4 years ago
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    seee, everytime i think i'm going crazy..but thank you guys, AGAIN.

  21. TuringTest
    • 4 years ago
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    very welcome!

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