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Thread: l' hospital rule

  1. #1
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    l' hospital rule

    Please help me to resolve this..

    limit (x-pi) (tan (x/2))
    x-pi

    I found the answer is 1/0.. i think its wrong...
    tq..
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  2. #2
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    Quote Originally Posted by maiya View Post
    Please help me to resolve this..

    limit (x-pi) (tan (x/2))
    x-pi

    I found the answer is 1/0.. i think its wrong...
    tq..
    \lim_{x \to \pi}\left[(x - \pi)\tan{\frac{x}{2}}\right] = \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right].

    This \to \frac{0}{0}, so you can use L'Hospital's Rule.

    \lim_{x \to \pi}\left[\frac{(x -  \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]

     = \frac{1}{-\frac{1}{2}}

     = -2.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    \lim_{x \to \pi}\left[(x - \pi)\tan{\frac{x}{2}}\right] = \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right].

    This \to \frac{0}{0}, so you can use L'Hospital's Rule.

    \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]

     = \frac{1}{-\frac{1}{2}}

     = -2.
    tqvm..
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  4. #4
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    Quote Originally Posted by Prove It View Post
    \lim_{x \to \pi}\left[(x - \pi)\tan{\frac{x}{2}}\right] = \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right].

    This \to \frac{0}{0}, so you can use L'Hospital's Rule.

    \lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]

     = \frac{1}{-\frac{1}{2}}

     = -2.
    sorry mr. prove it... can you show me the full way how to get the line 3? i can't get it..
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  5. #5
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    Quote Originally Posted by maiya View Post
    sorry mr. prove it... can you show me the full way how to get the line 3? i can't get it..
    Try substituting x = \pi...
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  6. #6
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    Do you mean
    <br />
\lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]<br />
?

    ProveIt thought you mean the third equation.

    L'Hopital's rule says that, if f(a)= g(a)= 0, then \lim_{x\to a} \frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}.

    In other words, he differentiated the numerator, (x- \pi)sin\left(\frac{x}{2}\right) and the denominator, cos\left(\frac{x}{2}\right), separately.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Do you mean
    <br />
\lim_{x \to \pi}\left[\frac{(x - \pi)\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right] = \lim_{x \to \pi}\left[\frac{\frac{1}{2}(x - \pi)\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{-\frac{1}{2}\sin{\frac{x}{2}}}\right]<br />
?

    ProveIt thought you mean the third equation.

    L'Hopital's rule says that, if f(a)= g(a)= 0, then \lim_{x\to a} \frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}.

    In other words, he differentiated the numerator, (x- \pi)sin\left(\frac{x}{2}\right) and the denominator, cos\left(\frac{x}{2}\right), separately.
    tqvm mr.prove it and HallsofIvy. I got it...
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