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Math Help - lets see if you can get this beast

  1. #1
    Junior Member winterwyrm's Avatar
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    lets see if you can get this beast

    lim(2-x)tan(πx/2)
    X->1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by winterwyrm View Post
    lim(2-x)tan(πx/2)
    X->1
    \lim_{x \to 1_+} (2-x) \tan(\pi x/2)=-\infty

    \lim_{x \to 1_-} (2-x) \tan(\pi x/2)=+\infty

    RonL
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  3. #3
    TD!
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    Seems rather trivial. Perhaps you mean the following?

    <br />
\mathop {\lim }\limits_{x \to 1} \left( {\left( {1 - x} \right)\tan \left( {\frac{{\pi x}}{2}} \right)} \right)<br />
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  4. #4
    Junior Member winterwyrm's Avatar
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    sorry, it's actually lim (2-x)^tan(Pi x/2)
    as x approaches 1

    I tried the formatting from word, and it didn't quite work
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by winterwyrm View Post
    sorry, it's actually lim (2-x)^tan(Pi x/2)
    as x approaches 1

    I tried the formatting from word, and it didn't quite work
    My calculator gives the answer as e^{2/\pi}, but I can't tell you why.

    -Dan
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  6. #6
    TD!
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    That's correct. Does this help you?

    <br />
\mathop {\lim }\limits_{x \to 1} \left( {2 - x} \right)^{\tan \left( {\frac{{\pi x}}{2}} \right)}  = \exp \left( {\mathop {\lim }\limits_{x \to 1} \left( {\tan \left( {\frac{{\pi x}}{2}} \right)\log \left( {2 - x} \right)} \right)} \right)<br />
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TD! View Post
    That's correct. Does this help you?

    <br />
\mathop {\lim }\limits_{x \to 1} \left( {2 - x} \right)^{\tan \left( {\frac{{\pi x}}{2}} \right)}  = \exp \left( {\mathop {\lim }\limits_{x \to 1} \left( {\tan \left( {\frac{{\pi x}}{2}} \right)\log \left( {2 - x} \right)} \right)} \right)<br />
    I always forget about that trick.

    -Dan
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  8. #8
    Junior Member winterwyrm's Avatar
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    isn't tan(Pi x/2) indeterminate though? I've tried tons of tricks, and I can't seem to crack any way to get the pi x/2 out of there.
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  9. #9
    TD!
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    <br />
\mathop {\lim }\limits_{x \to 1} \left( {\tan \left( {\frac{{\pi x}}{2}} \right)\log \left( {2 - x} \right)} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\log \left( {2 - x} \right)}}{{\cot \left( {\frac{{\pi x}}{2}} \right)}}<br />

    Now you can apply l'H˘pital's rule.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by winterwyrm View Post
    isn't tan(Pi x/2) indeterminate though? I've tried tons of tricks, and I can't seem to crack any way to get the pi x/2 out of there.
    Yes, but it's an indeterminate of the form \infty \cdot 0. Do you know how to handle that?

    -Dan
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  11. #11
    Junior Member winterwyrm's Avatar
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    whoa, you guys are awesome, but do you have to handle different indet forms differently? if so, do you just tackle it by taking the derivative? I derivatived it before using the ln/exp rule, every time, lol. Thanks a million to everyone.
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  12. #12
    TD!
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    You can only apply l'H˘pital's rule (so f/g -> f'/g' for the limit) to the indeterminate forms 0/0 or ∞/∞. If you have another indeterminate form (0.∞, 1^∞, ...), you have to reduce it to one of the "allowed" indeterminate forms for l'H˘pital's rule.
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by winterwyrm View Post
    sorry, it's actually lim (2-x)^tan(Pi x/2)
    as x approaches 1

    I tried the formatting from word, and it didn't quite work
    We are interested in:

    <br />
\lim_{x \to 1} ~(2-x)^ { \tan(\pi x/2)}<br />

    Put y=x-1, then:

    <br />
~(2-x)^{ \tan(\pi x/2)}=(1-y)^{\cot(\pi y/2)}<br />

    but \cot(\pi y/2) \approx 2/(\pi y) for small y

    So as y \to 0, \ \ (1-y)^{\cot(\pi y/2)} \to (1-y)^{\frac{2}{\pi y}}

    and with some jiggery pokery this last should come to e^{2/pi} (it should rearrange into the classical limit form for an exponential)

    RonL
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TD! View Post
    You can only apply l'H˘pital's rule (so f/g -> f'/g' for the limit) to the indeterminate forms 0/0 or ∞/∞. If you have another indeterminate form (0.∞, 1^∞, ...), you have to reduce it to one of the "allowed" indeterminate forms for l'H˘pital's rule.
    Such as \lim_{x \to 1} \frac{tan(\pi x/2)}{\frac{1}{ln(2 - x)}}...

    -Dan
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  15. #15
    TD!
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    Yes, although I prefer the way of rewriting as I did (log(...)/cot(...)) because log is easier to differentiate than 1/log.
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