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Math Help - Limit using L'H˘pital

  1. #1
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    Limit using L'H˘pital

    Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

    \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}  {x^2}}}

    I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

    \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}  {x^2}}

    And then:

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}

    How should I proceed?

    I've tried using L'H˘pitals rule, but I didn't get too far.

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}

    From here it would seem to be that the answer is e^{-\infty} (which I think it would be equal to zero) but I used derive to make the calculus, and the lÝmit should give e^{-1/6}

    Bye there.
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    Quote Originally Posted by Ulysses View Post
    Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

    \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}  {x^2}}}

    I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

    \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}  {x^2}}

    And then:

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}

    How should I proceed?

    I've tried using L'H˘pitals rule, but I didn't get too far.

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}

    Bye there.


    \left(\ln\frac{\sin x}{x}\right)^{'}=\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2} =\frac{x\cos x-\sin x}{x\sin x} , so applying L'Hospital you get:

    \frac{x\cos x-\sin x}{2x^2\sin x} , and L'H once again: \frac{-x\sin x}{4x\sin x+2x^2\cos x} , and once again: \frac{-\sin x-x\cos x}{4\sin x+8x\cos x-2x^2\sin x} , and again:

    \frac{-2\cos x+x\sin x}{12\cos x-12x\sin x-2x^2\cos x}\xrightarrow [x\to o^+]{}-\frac{2}{12}=-\frac{1}{6} , so the limit is e^{-1/6}

    Tonio
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  3. #3
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    Thank you very much Tonio!
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  4. #4
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    Quote Originally Posted by Ulysses View Post
    Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

    \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}  {x^2}}}

    I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

    \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}  {x^2}}

    And then:

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}

    How should I proceed?

    Bye there.
    \lim_{x \to 0^{+}}\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\rig  ht] = \lim_{x \to 0^{+}}e^{\ln{\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\rig  ht]}}

     = \lim_{x \to 0^{+}}e^{\frac{\ln{\left(\frac{\sin{x}}{x}\right)}  }{x^2}}

     = e^{\lim_{x \to 0^{+}}\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x  ^2}}

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

    e^{\lim_{x \to 0^{+}}\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x  ^2}} = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{\frac{x\cos{x} - \sin{x}}{x^2}}{\frac{\sin{x}}{x}}}{2x}\right)}

     = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{x\cos{x} - \sin{x}}{x\sin{x}}}{2x}\right)}

     = e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)}.

    This still \to \frac{0}{0} so use L'Hospital's Rule again...

    e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} -  \sin{x}}{2x^2\sin{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(\frac{\cos{x} - x\sin{x} - \cos{x}}{2x^2\cos{x} + 4x\sin{x}}\right)}

     = e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)}

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

    e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(-\frac{\cos{x}}{4\cos{x} + 2x\sin{x} + 2\cos{x}}\right)}

     = e^{-\frac{1}{4 + 2}}

     = e^{-\frac{1}{6}}.



    PHEW!

    Edit: Fixed a small mistake.
    Last edited by Prove It; May 24th 2010 at 08:26 PM.
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  5. #5
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    Quote Originally Posted by tonio View Post
    \left(\ln\frac{\sin x}{x}\right)^{'}=\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2} =\frac{x\cos x-\sin x}{x\sin x} , so applying L'Hospital you get:

    \frac{x\cos x-\sin x}{2x^2\sin x} , and L'H once again: \frac{-x\sin x}{4x\sin x+2x^2\cos x} , and once again: \frac{-\sin x-x\cos x}{4\sin x+8x\cos x-2x^2\sin x} , and again:

    \frac{-2\cos x+x\sin x}{12\cos x-12x\sin x-2x^2\cos x}\xrightarrow [x\to o^+]{}-\frac{2}{12}=-\frac{1}{6} , so the limit is e^{-1/6}

    Tonio
    You've forgotten about the 2x in the denominator that comes from the power in the logarithm.
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  6. #6
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    In general to get the  1^\infty indeterminate form into the form  \frac00 do the following:

    If  \lim_{x\to a}f(x) = 1 and  \lim_{x\to a}g(x) = \infty

    then  \lim_{x\to a} f(x)^{g(x)} = \exp\left(\lim_{x\to a}\frac{\ln f(x)}{1/g(x)}\right) .

    Then just apply L'H˘pital's rule to get your answer.
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  7. #7
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    I wanted to ask him about that Prove it. Thanks. And I know that the result that he get to its good, because its the same that Derive gave it to me.

    I have to go now, but I'll read this later.

    Bye there, and thanks to both of you.
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    Quote Originally Posted by Ulysses View Post
    Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

    \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1}  {x^2}}}

    I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

    \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})}  {x^2}}

    And then:

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}

    How should I proceed?

    I've tried using L'H˘pitals rule, but I didn't get too far.

    \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}

    From here it would seem to be that the answer is e^{-\infty} (which I think it would be equal to zero) but I used derive to make the calculus, and the lÝmit should give e^{-1/6}

    Bye there.

    We have  \frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ...


     \lim_{x\to 0} \left(  \frac{\sin(x)}{x}  \right)^{\frac{1}{x^2}} = \lim_{x\to 0} \left(  1 - \frac{x^2}{6} + \frac{x^4}{120} - ...\right)^{\frac{1}{x^2}}

    Let  n = \frac{1}{x^2} and we have

     \lim_{n \to \infty} \left( 1 - \frac{1}{n}( \frac{1}{6} - \frac{1}{120 n} - ...) \right )^n

     = exp \{\ -(  \lim_{n \to \infty} \frac{1}{6} - \frac{1}{120 n} - ... ) \}\

     = e^{-1/6}

    EDIT: I think it is unnecessary to evaluate only the right hand limit .
    Last edited by simplependulum; May 24th 2010 at 08:08 PM.
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    <br />
e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(\frac{\cos{x} - x\sin{x} - \cos{x}}{2x^2\cos{x} - 4x\sin{x}}\right)}<br />
    Mistake appears in this line

    The derivative of  2x^2 \sin(x) is

     4x\sin(x) + 2x^2 \cos(x)
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  10. #10
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    Quote Originally Posted by simplependulum View Post
    Mistake appears in this line

    The derivative of  2x^2 \sin(x) is

     4x\sin(x) + 2x^2 \cos(x)
    Thanks.
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    Quote Originally Posted by Prove It View Post
    You've forgotten about the 2x in the denominator that comes from the power in the logarithm.

    Where exactly? I don't think I'd gottent the correct answer had I forgotten that, but perhaps I did. Where?

    Tonio
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    Well, the result that he gets its the right one.

    It's great to have so many focus for the same problem. Thanks to all of you.

    I wanted to ask about this one in particullar:

    Quote Originally Posted by simplependulum View Post
    We have  \frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - ...


     \lim_{x\to 0} \left( \frac{\sin(x)}{x} \right)^{\frac{1}{x^2}} = \lim_{x\to 0} \left( 1 - \frac{x^2}{6} + \frac{x^4}{120} - ...\right)^{\frac{1}{x^2}}

    Let  n = \frac{1}{x^2} and we have

     \lim_{n \to \infty} \left( 1 - \frac{1}{n}( \frac{1}{6} - \frac{1}{120 n} - ...) \right )^n

     = exp \{\ -(  \lim_{n \to \infty} \frac{1}{6} - \frac{1}{120 n} - ... ) \}\

     = e^{-1/6}

    EDIT: I think it is unnecessary to evaluate only the right hand limit .
    Are you using Taylors polinomial there? I haven't seen it yet, but its good to know it can be solved that way.

    Bye there.

    EDIT: Now that I've taken a deeper look on tonios reasoning i've seen that he proceeded right and didn't forget a thing. It's just that as he started by reasoning the logaritm part I thought that he forgot the denominator, but now I see that he didn't.
    Last edited by Ulysses; May 25th 2010 at 12:01 PM.
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