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Math Help - How to calculate this limit?

  1. #1
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    How to calculate this limit?

    Hello forum members, thanks for all the help so far,
    I would like to know how to calculate the limit below.

    \lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}

    As you can see I've finally understood how to write equations using the math tool.

    Thanks for the help.
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  2. #2
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    Smile

    Quote Originally Posted by azarue View Post
    Hello forum members, thanks for all the help so far,
    I would like to know how to calculate the limit below.

    \lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}

    As you can see I've finally understood how to write equations using the math tool.

    Thanks for the help.
    i believe you have \lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}=0 since the numerator is zero and the denominator is not.
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  3. #3
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    I don't think so.

    Because X ---> 1+

    I can't calculate a limit with the form '0/0'.
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  4. #4
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    Quote Originally Posted by Raoh View Post
    i believe you have \lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}}=0 since the numerator is zero and the denominator is not.
    \lim_{x \to 1^+}1-x = 0

    OP: Do you know about L'Hopital's rule?
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  5. #5
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    I know this rule, I just can't use it.

    I went to an online limit calculator and it shows that :

    \lim_{x\to1^+}{{55\,\sin \left(\pi\,x\right)}\over{1-x}} = 55\,\pi

    How can I get this result without using L'L'Hopital's rule?
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  6. #6
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    Quote Originally Posted by azarue View Post
    Because X ---> 1+

    I can't calculate a limit with the form '0/0'.
    sorry what i wrote is wrong
    put h=1-x you'll find \lim_{h\to 0}{{55\,\sin \left(\pi\,(1-h)\right)}\over{h}}=55\pi
    here's how
    \lim_{h\to 0}{{55\,\sin \left(\pi\,(1-h)\right)}\over{h}}= 55 \lim_{ h \rightarrow 0}\frac{\sin(\pi-\pi h) }{ h} = 55\lim_{ h \rightarrow 0}-\pi \frac{\sin(-\pi h) }{h \pi}=55\pi
    Last edited by Raoh; May 3rd 2010 at 06:43 AM.
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  7. #7
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    thanks for the help

    that was helpful. thank you.
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