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Math Help - Limit question

  1. #1
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    Limit question

    I am wondering if there is a way to calculate limits or functions as x=> a certain point say x=>1 using derivatives? I remember there being a way to do it but I can't find any stuff online or in books

    thanks
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  2. #2
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    Hello,
    Quote Originally Posted by Calky View Post
    I am wondering if there is a way to calculate limits or functions as x=> a certain point say x=>1 using derivatives? I remember there being a way to do it but I can't find any stuff online or in books

    thanks
    l'H˘pital's rule - Wikipedia, the free encyclopedia



    But it's not for any sort of limits. Sometimes (and often), you'll have to think by yourself

    -------------------------------
    Okay, there may be something else (since I'm not sure what you're asking for ^^') !

    \lim_{x \to a} \frac{f(x)-f(a)}{x-a}=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}=f ~'(a)
    Last edited by Moo; August 31st 2008 at 09:09 AM.
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  3. #3
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    Sometimes you might need to use the definition of the derivative:
    \lim_{x \to a} \frac{f(x) - f(a)}{x - a}

    To compute limits like: \lim_{h \to 0} \frac{5^h - 2^h}{h}
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  4. #4
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    so this would be correct?




    I don't know why but I always find wikipedia definitions and explanations hard to comprehend
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  5. #5
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    The exponent is 1/2 not -1/2. It's already in the denominator.

    EDIT: or is it?

    EDIT2: You are still incorrect.
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  6. #6
    Moo
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    Quote Originally Posted by Calky View Post
    so this would be correct?




    I don't know why but I always find wikipedia definitions and explanations hard to comprehend
    Do you know about conjugates ?

    This is mostly about this identity : (a-b)(a+b)=a^2-b^2

    So when you have a square root that's disturbing you, you can multiply by its conjugate, because it will square the square root.

    For example, the conjugate of 3-\sqrt{x} is 3+\sqrt{x}

    So the trick (that you can use in many situations) is to multiply by 1=\frac{3+\sqrt{x}}{3+\sqrt{x}}

    ------------------------------------
    Otherwise, your derivative (not the answer) is correct.
    Remember that l'Hospital's rule is used in situations \frac \infty \infty or \frac 00
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  7. #7
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    I found that the derivative of 3-x^(1/2) = -1/2x^(-1/2) using the power rule is that incorrect? is that what you are saying
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  8. #8
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    Quote Originally Posted by Moo View Post
    Do you know about conjugates ?

    This is mostly about this identity : (a-b)(a+b)=a^2-b^2

    So when you have a square root that's disturbing you, you can multiply by its conjugate, because it will square the square root.

    For example, the conjugate of 3-\sqrt{x} is 3+\sqrt{x}

    So the trick (that you can use in many situations) is to multiply by 1=\frac{3+\sqrt{x}}{3+\sqrt{x}}

    ------------------------------------
    Otherwise, your derivative (not the answer) is correct.
    Remember that l'Hospital's rule is used in situations \frac \infty \infty or \frac 00

    ok thank you

    I did it that way and found that you would be dividing by zero.......
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  9. #9
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    If you meant:
    -\frac{1}{2} x^{-1/2}

    Then you are correct.
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  10. #10
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    Quote Originally Posted by Chop Suey View Post
    If you meant:
    -\frac{1}{2} x^{-1/2}

    Then you are correct.

    yes that is what I meant. maybe my notation or handwriting is a little off



    thanks a lot
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  11. #11
    Moo
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    Quote Originally Posted by Calky View Post
    ok thank you

    I did it that way and found that you would be dividing by zero.......
    No, because in the denominator, you'll get something that will simplify with the numerator
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  12. #12
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    Quote Originally Posted by Moo View Post
    No, because in the denominator, you'll get something that will simplify with the numerator

    oh ok I see that now, I ended up doing it the other way


    can you use l'hopital rule for piecewise functions?

    for example with

    f(x) = { 2x-6/x-3, x =/= 3
    ..........{ 5, x=3


    finding limit of f(x) as x=> 3
    Last edited by Calky; August 31st 2008 at 10:05 AM.
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