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Math Help - Limit Problems

  1. #1
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    Limit Problems

    These are the only two i couldn't figure out. If someone could help, that'd be great. Here they are:

    1) Limit where h approaches 0- {Square root(4 + h) - 2}/h

    In words, the limit where h approaches zero of the square root of four plus h minus 2 all over h.

    2) Limit where h approaches 0- 1/{square root(4 + h)} - 1/{2}/h

    In words, the limit where h approaches zero of 1 divided by the square root of four plus h minus one half all over h.


    Please help, need by tomorrow morning! Thanks!
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  2. #2
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    1) lim ((squ(4+h)-2)/h)((squ(4+h)+2)/(squ(4+h)+2) =

    lim (4+h-4)/(h*squ(4+h)+2) =

    lim 1/(squ(4+h)+2 =

    1/(2+2) =

    1/4

    2)

    0

    I don't know if i understood part 2 correctly...
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  3. #3
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    Ok, well thank you for the first one.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Blue Griffin View Post
    2) Limit where h approaches 0- 1/{square root(4 + h)} - 1/{2}/h
    well, what are your options here? try to simplify the top so that we cancel the h in the bottom, that way we can take the limit. how do we simplify the top? well, we can start by combining the fractions.

    \lim_{h \to 0} \frac {\frac 1{\sqrt {4 + h}} - \frac 12}h

    = \lim_{h \to 0} \frac {\frac {2 - \sqrt{4 + h}}{2 \sqrt{4 + h}}}{h}

    \lim_{h \to 0} \frac {2 - \sqrt{4 + h}}{2h \sqrt {4 + h}}

    well, now what? the h doesn't seem to cancel, so now we apply a common trick, conjugation of the numerator...

    \Rightarrow \lim_{h \to 0} \frac {2 - \sqrt{4 + h}}{2h \sqrt {4 + h}} \cdot \frac {2 + \sqrt{4 + h}}{2 + \sqrt {4 + h}}

    now continue and see where you get with this. your answer should be - \frac 1{16}
    Last edited by Jhevon; October 5th 2007 at 01:59 PM.
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  5. #5
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    Yeah definitely didn't read the second one correctly
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  6. #6
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    Thanks a lot guys, i understand it now, and i finished my homework!
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