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Math Help - Stuck on a trig limit

  1. #1
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    Arrow Stuck on a trig limit

    \lim_{x\to\\1^+}\frac{x+1}{x\sin{{\pi}x}}

    I was never any good at these. >.<
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    Quote Originally Posted by xxlvh View Post
    \lim_{x\to\\1^+}\frac{x+1}{x\sin{{\pi}x}}

    I was never any good at these. >.<
    Surely since \sin{\pi x} \to 0 as x \to 1, and x + 1 \to 2, then

    \frac{x + 1}{x\sin{\pi x}} \to \frac{2}{0} \to \infty...
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    Is there no possible way to rearrange it? I was told to always try that when a denominator went to 0, which of course I can't seem to do on this question..
    Also the answer is negative infinity, not positive.
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    Quote Originally Posted by xxlvh View Post
    Is there no possible way to rearrange it? I was told to always try that when a denominator went to 0, which of course I can't seem to do on this question..
    Also the answer is negative infinity, not positive.
    I didn't notice the "right hand limit" part.

    There are several ways to rearrange the expression, including

    \frac{x + 1}{x \sin{\pi x}} = \frac{x}{x\sin{\pi x}} + \frac{1}{x\sin{\pi x}}

     = \frac{1}{\sin{\pi x}} + \frac{1}{x \sin{\pi x}}

     = \frac{1}{\sin{\pi x}}\left(1 + \frac{1}{x}\right).


    All will lead to the same conclusion though.
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  5. #5
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    You could do l'Hospital's rule and get: \frac{1}{x\pi\cos(\pi x)+\sin(\pi x)} \rightarrow -\frac{1}{\pi}
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    I don't believe L'Hopitals is applicable in this case, the numberator does not go to 0.

    Hmm, is there any way of obtaining the answer of negative infinity algebraically then?
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    Quote Originally Posted by courteous View Post
    You could do l'Hospital's rule and get: \frac{1}{x\pi\cos(\pi x)+\sin(\pi x)} \rightarrow -\frac{1}{\pi}
    You most definitely can NOT use L'Hospital's Rule in this case.

    To use L'Hospital, the expression must tend to an indefinite form, i.e. \frac{0}{0} or \frac{\infty}{\infty}.

    But like I showed earlier, the expression tends to \frac{2}{0}.
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  8. #8
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    Quote Originally Posted by xxlvh View Post
    [snip]Also the answer is negative infinity, not positive.
    Note that \lim_{x \rightarrow 1^+} \sin (\pi x) = 0^- since the approach is from the third quadrant.
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