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Math Help - L'Hospital's rule

  1. #1
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    L'Hospital's rule

    I'm having a difficulty differentiating this using L'Hospital's rule:

    lim x->(infinite) xsin(pi/x)

    Applying L'Hospital's rule and differentiating I got:

    lim x->(infinite) xcos(pi/x) + sin(pi/x)

    and that equals = (infinite)(1) + 0 = infinite

    The correct answer in my textbook is just "pi"

    Can someone tell me what I'm doing wrong?
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  2. #2
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    If your function is: x*sin\frac{\pi}{x} then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.
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  3. #3
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    Well, to use L'Hospital you must have an indefinite form. Second you differentiate each part seperately. Make sure you check what the indefinite forms are.
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  4. #4
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    Quote Originally Posted by Alterah View Post
    If your function is: x*sin\frac{\pi}{x} then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.
    Well, then can you explain why the answer is pi?
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  5. #5
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    But the function is of indeterminate form. I.e,  0*\infty so we can rewrite the limit so it is of the form  \frac{0}{0}

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

    we see \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} = \frac{0}{0}

    apply L'Hospital's Rule
    \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} =<br />
\lim_{x\rightarrow\infty} \frac{\frac{-\pi}{x^{2}}\cos{\frac{\pi}{x}}}{\frac{-1}{x^{2}}} = \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}}

    Since  \lim_{x\rightarrow\infty} \frac{1}{x} = 0

    \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi
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  6. #6
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    Haven, you beat me to it. Took me a second to realize that it was actually an indeterminate form.
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  7. #7
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    Quote Originally Posted by Haven View Post
    But the function is of indeterminate form. I.e,  0*\infty so we can rewrite the limit so it is of the form  \frac{0}{0}

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

    we see \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} = \frac{0}{0}

    apply L'Hospital's Rule
    \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} =<br />
\lim_{x\rightarrow\infty} \frac{\frac{-\pi}{x^{2}}\cos{\frac{\pi}{x}}}{\frac{-1}{x^{2}}} = \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}}

    Since  \lim_{x\rightarrow\infty} \frac{1}{x} = 0

    \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi
    The problem wasn't but could be transformed into indeterminate form. I was hoping the OP would see how to transform it, and then solve the problem.
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  8. #8
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    Quote Originally Posted by lvleph View Post
    The problem wasn't but could be transformed into indeterminate form. I was hoping the OP would see how to transform it, and then solve the problem.
    Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.
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  9. #9
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    Quote Originally Posted by Eternal View Post
    Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.
    Notice that the problem is exactly the same, but has been manipulated so that it looks different and you get one of the indeterminate forms.
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  10. #10
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    Another question I'm confused on:

    lim x->0+ Sin(x)lnx

    How would I solve this? ln(0) is undefined...
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  11. #11
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    Make the substitution \ln x = u. Then we have \lim_{u \to -\infty}\sin (e^{u}) u.
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