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Math Help - Evaluating Limits

  1. #1
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    Evaluating Limits

    i'm so lost on this problem.. I don't even know where to start on this one. Help me guys

    Evaluate.

    Lim (3+h)^-1 - 3^-1 / h
    h->0
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  2. #2
    Super Member 11rdc11's Avatar
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    \lim_{h \to 0} \frac{1}{3 + h} - 3^{-\frac{1}{h}}

    Does it look like this?
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  3. #3
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    3 ways:
    - Combine fractions in the numerator
    - Multiply by 3(3+h) on both sides of the fraction
    - Treat it as a derivative and use power/quotient rule to evaluate it at h = 0.
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  4. #4
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    Quote Originally Posted by 11rdc11 View Post
    \lim_{h \to 0} \frac{1}{3 + h} - 3^{-\frac{1}{h}}

    Does it look like this?
    It looks like he meant:
    \lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}

    I could be wrong though.
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Chop Suey View Post
    It looks like he meant:
    \lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}

    I could be wrong though.
    Hopefully becuase the other limit a little bit harder to do.
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  6. #6
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    Hey guys,

    Sorry to bump this thread up again but i've been trying to figure the answer out and keep getting 0/9, which is incorrect. Any help on pointing me in the right direction? Thanks everybody.
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  7. #7
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    \lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}

    Oh and the problem was this one
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  8. #8
    Super Member 11rdc11's Avatar
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    \lim_{h \to 0} \frac{\displaystyle\frac{1}{3+h} - \frac{1}{3}}{h}

    ok like chop suey said find a common denominator

    \lim_{h \to 0}\frac{\frac{-h}{9+3h}}{h}

    \lim_{h \to 0} \frac{-h}{h(9+3h)}

    Can you finish up from here?
    Last edited by 11rdc11; September 9th 2008 at 03:02 PM. Reason: Made correction
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  9. #9
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    the common denominator would be (3+h)(3) right? So I multiply that top and bottom and I get this

    3 - 3 - h / h (3+h)(3)

    am I doing something wrong?
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  10. #10
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    but if you factored out a 3 wouldn't it be 2 + h ?
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  11. #11
    Super Member 11rdc11's Avatar
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    Yep your right I made a mistake. Thanks for pointing that out. The right should be -\frac{1}{9}
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  12. #12
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    Thanks alot man you really helped me understand the concept .
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  13. #13
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    Sorry to bump this thread again but I had another question pertaining to the problem from before.

    The slope would equal -1/9 right? My professor had this answer down.

    "The slope of the tangent line to y = 1/x at the point (3, 1/3) is -1/9."

    How did he get the points (3,1/3) and y = 1/x?
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  14. #14
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    The derivative is the slope of the tangent line at a point x=a. The definition is:
    \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

    This is very similar to your limit. The given question had already plugged in the point, but one can find the original function easily. Let's revert 3 back to x:

    f(x+h) = \frac{1}{x+h}

    f(x) = \frac{1}{x}

    f(3) = \frac{1}{3}

    So -\frac{1}{9} is the slope of the tangent line to point (3, \frac{1}{3})
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  15. #15
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    So basically f(x) = 1/x was the original equation? All I had to do was plug the X value which is 3 back into the original equation to find my Y value?
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