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Math Help - limit using l'hopital

  1. #1
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    Question limit using l'hopital

    this is my question:

    lim [tan(x)/(x)]/(1/(x^2)
    x->0

    I have tried to work this out and I know how to get to: [ ln(tan(x)/x) ]/x^2
    but then when I start using l'hopital I get stuck.
    PLEASE DO NOT TELL ME THAT IT'S e^1/3, because I want to see the way to solve this and not the final answer!

    THANKS!
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  2. #2
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    \lim_{x\rightarrow{0}}\frac{\frac{tan(x)}{x}}{\fra  c{1}{x^{2}}}=\lim_{x\rightarrow{0}}[xtan(x)]

    Are you sure you posted it correctly. Not much need for L'Hopital. The limit is merely 0.
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  3. #3
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    Quote Originally Posted by Lilly100 View Post
    this is my question:

    lim [tan(x)/(x)]/(1/(x^2)
    x->0

    I have tried to work this out and I know how to get to: [ ln(tan(x)/x) ]/x^2
    but then when I start using l'hopital I get stuck.
    PLEASE DO NOT TELL ME THAT IT'S e^1/3, because I want to see the way to solve this and not the final answer!

    THANKS!
    Quote Originally Posted by galactus View Post
    \lim_{x\rightarrow{0}}\frac{\frac{tan(x)}{x}}{\fra  c{1}{x^{2}}}=\lim_{x\rightarrow{0}}[xtan(x)]

    Are you sure you posted it correctly. Not much need for L'Hopital. The limit is merely 0.
    Not to mention that when you apply L'Hopital's rule to this problem I get that
    \lim_{x \to 0}\frac{\frac{tan(x)}{x}}{\frac{1}{x^{2}}} = \lim_{x \to 0} \frac{x(sin(x)~cos(x) - x)}{2~cos^2(x)}
    not what you wrote.

    -Dan
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  4. #4
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    what I posted

    This is the original limit:

    lim [tan(x)/(x)]/(1/(x^2)
    x->0

    it reads:
    tan(x)/(x) is the base
    1/(x^2) is the power

    galactus:
    notice that tanx/x is in ln form
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  5. #5
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    It would've helped if you would have told us 1/x^2 was a power in the first place.

    Sorry for the small font. I am unable to make it larger.

    Rewrite as:

    e^{\lim_{x\rightarrow{0}}\frac{ln(\frac{tan(x)}{x}  )}{x^{2}}}

    Using L'Hopital:

    e^{\frac{1}{2}\lim_{x\rightarrow{0}}\frac{x+xtan^{  2}x-tanx}{x^{2}tanx}}

    L'Hopital again:

    e^{\lim_{x\rightarrow{0}}\frac{tanx}{2tanx+x+xtan^  {2}x}\cdot\lim_{x\rightarrow{0}}[1+tan^{2}x]}

    The limit to the right is just 1 so we can shed it and use L'Hopital again:

    e^{\lim_{x\rightarrow{0}}\frac{1}{2xtanx+3}}

    Of course, after all this we should be able to see the limit is 1/3 and we have:

    e^{\frac{1}{3}}

    You could also break it up and attempt another route.

    \frac{tan(x)}{x}=\frac{1}{cos(x)}\cdot\frac{sin(x)  }{x}

    e^{\frac{ln(\frac{1}{cos(x)})\cdot\frac{sin(x)}{x}  )}{x^{2}}}

    e^{\frac{ln(sec(x))}{x^{2}}}\cdot{e^{\frac{ln(\fra  c{sin(x)}{x})}{x^{2}}}}

    The left limit is \sqrt{e} and the right one is e^{\frac{-1}{6}}

    e^{\frac{1}{2}}\cdot{e^{\frac{-1}{6}}}=e^{\frac{1}{3}}

    There are many ways to go about it. The method WITHOUT L'Hopital is the most interesting.
    Last edited by galactus; December 22nd 2007 at 03:45 PM.
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  6. #6
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    galactus!

    Thank you for solving this.

    If you could explain the first l'hopital that you did, that would be great!

    when I did l'hopital for this-

    ln[tan(x)/(x)]/x^2

    i got to:
    for the numerator {1/ [tan(x)/(x)]*[1/(cosx)^2]*x-[tan(x)*1]} / x^2
    for the denominator 2x

    what am I doing wrong?

    thanks again
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  7. #7
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    I just took the derivative of ln(\frac{tan(x)}{x})

    Which is \frac{x+xtan^{2}(x)-tan(x)}{x^{2}tan(x)}

    Chain rule:

    Derivative of outside:

    \frac{1}{\frac{tan(x)}{x}}=\frac{x}{tan(x)}

    Derivative of inside:

    \frac{d}{dx}[\frac{tan(x)}{x}]=\frac{x(1+tan^{2}(x))-tan(x)}{x^{2}}

    \frac{x}{tan(x)}\cdot\frac{x(1+tan^{2}(x))-tan(x)}{x^{2}}

    = \frac{x+xtan^{2}(x)-tan(x)}{xtan(x)}
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