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Math Help - L'hospital's rule problem

  1. #1
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    L'hospital's rule problem

    Hi guys I'm having trouble understanding this concept on problems like these

    1. Lim CosX / 1 - Sinx
    x->(pi/2)+
    the answer would be Negative Infinity



    2. Lim e^x - 1 / x^3
    x->0
    the answer would be Infinity

    My question is why would the answers be infinity or negative infinity. I have no clue what is going on. Any detailed, step by step help would be greatly appreciated. Thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Afterme View Post
    Hi guys I'm having trouble understanding this concept on problems like these

    1. Lim CosX / 1 - Sinx
    x->(pi/2)+
    the answer would be Negative Infinity



    2. Lim e^x - 1 / x^3
    x->0
    the answer would be Infinity

    My question is why would the answers be infinity or negative infinity. I have no clue what is going on. Any detailed, step by step help would be greatly appreciated. Thanks
    For the first one Rewrite it as

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

    Now let t=\frac{\pi}{2}-x

    So as x\to\frac{\pi}{2}\Rightarrow{t\to{0}}

    So we have

    \lim_{t\to{0}}\frac{\sin(t)}{1-\cos(t)}

    Now \sin(x)\sim{x}\text{ As }x\to{0}

    and

    1-\cos(x)\sim\frac{x^2}{2}\text{ As }x\to{0}

    So we can say that

    \lim_{t\to{0}}\frac{\sin(t)}{1-\cos(t)}\sim\lim_{t\to{0}}\frac{t}{\frac{t^2}{2}}=  \lim_{t\to{0}}\frac{2}{t}

    Now can you see why?

    Or if you wanted to do L'hopital's

    Since \sin(0)=1-\cos(0)=0

    we have that

    \lim_{t\to{0}}\frac{\sin(t)}{1-\cos(t)}=\lim_{t\to{0}}\frac{\cos(t)}{-\sin(t)}

    Now we cannot apply L'hopital's again since it is not \frac{0}{0} but \frac{1}{0}
    ---------------------------------------------------------------------------

    Now for the second one

    e^x-1\sim{x}\text{ As }x\to{0}

    So we have that


    \lim_{x\to{0}}\frac{e^x-1}{x^3}\sim\lim_{x\to{0}}\frac{x}{x^3}=\lim_{x\to{  0}}\frac{1}{x^2}

    Now you should be able to see whats the deal there.


    Or with L'hopitals

    \lim_{x\to{0}}\frac{e^x-1}{x^3}=\lim_{x\to{0}}\frac{e^x}{3x^2}

    Now we cannot apply Lhopitals again since this is \frac{1}{0}
    Last edited by Mathstud28; July 16th 2008 at 03:24 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Afterme View Post
    1. Lim CosX / 1 - Sinx
    x->(pi/2)+
    the answer would be Negative Infinity
    \lim_{x \to \frac{\pi }{2} ^2} \frac{cos(x)}{1 - sin(x)}

    This is of the form 0/0 so we can apply L'Hopital's rule.
    = \lim_{x \to \frac{\pi }{2} ^2} \frac{-sin(x)}{cos(x)}

    This is of the form -1/0, so the limit is at negative infinity.

    -Dan
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by topsquark View Post
    \lim_{x \to \frac{\pi }{2} ^2} \frac{cos(x)}{1 - sin(x)}

    This is of the form 0/0 so we can apply L'Hopital's rule.
    = \lim_{x \to \frac{\pi }{2} ^2} \frac{-sin(x)}{cos(x)}

    This is of the form -1/0, so the limit is at negative infinity.

    -Dan
    I am sorry, just out of curisoity, was there something wrong with my work?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Afterme View Post
    2. Lim e^x - 1 / x^3
    x->0
    the answer would be Infinity
    This is of the form 0/0, so using L'Hopital's rule:
    \lim_{x \to 0} \frac{e^x - 1}{x^3} = \lim_{x \to 0} \frac{e^x}{3x^2}

    This is of the form 1/0, so it is positive infinity.

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    \lim_{x \to \frac{\pi }{2} ^2} \frac{cos(x)}{1 - sin(x)}

    This is of the form 0/0 so we can apply L'Hopital's rule.
    = \lim_{x \to \frac{\pi }{2} ^2} \frac{-sin(x)}{cos(x)}

    This is of the form -1/0, so the limit is at negative infinity.

    -Dan

    Hey Dan, thanks for the help but how does -1/0 become negative infinity? I thought -1/0 was an indeterminate answer.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Afterme View Post
    Hey Dan, thanks for the help but how does -1/0 become negative infinity? I thought -1/0 was an indeterminate answer.
    Think about it numerically

    \frac{-1}{.00000000\cdots{1}}=?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Afterme View Post
    Hey Dan, thanks for the help but how does -1/0 become negative infinity? I thought -1/0 was an indeterminate answer.
    Quote Originally Posted by Mathstud28 View Post
    Think about it numerically

    \frac{-1}{.00000000\cdots{1}}=?
    I'll back that up with the additional comment that "infinity" itself is an indeterminant answer.

    -Dan
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  9. #9
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    It's because it's not really -1/0

    Rather, it is -1/x where x is continuously approaching 0 but never reaching it.

    Such number is hence continuously getting more negative, only bounded by the limit of negative infinity.
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