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Math Help - I don't understand this limit problem

  1. #1
    Member Jones's Avatar
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    I don't understand this limit problem

    Hi,


    I have
     \lim_{x \to 0+} ~\frac{sin^2x}{tan~x-x}

    Now replace the trig functions with its corresponding taylor polynomial

    \frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg  )-x}


    This limit evaluates to infinity, i don't understand why. It looks more like -2 to me
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  2. #2
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    Too much work (unless you are required to use Taylor representations), both functions approach zero, so you can easily just use L'hopitals rule.
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  3. #3
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    Quote Originally Posted by Jones View Post
    Hi,


    I have
     \lim_{x \to 0+} ~\frac{sin^2x}{tan~x-x}

    Now replace the trig functions with its corresponding taylor polynomial

    \frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg  )-x}


    This limit evaluates to infinity, i don't understand why. It looks more like -2 to me

    \frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg  )-x}= \frac{x^2-\frac{x^4}{6}+\frac{x^6}{36}}{\frac{x^3}{3}}=\frac  {3}{x}-\frac{x}{2}+\frac{x^3}{12}\xrightarrow [x\to 0^+]{}\infty

    Tonio
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  4. #4
    MHF Contributor matheagle's Avatar
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    or you can use L'Hopital's Rule here, which I just noticed was also recommended.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jones View Post
    Hi,


    I have
     \lim_{x \to 0+} ~\frac{sin^2x}{tan~x-x}

    Now replace the trig functions with its corresponding taylor polynomial

    \frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg  )-x}


    This limit evaluates to infinity, i don't understand why. It looks more like -2 to me
    Another convoluted way to show the limit doesn't exist is to assume it does. Then \lim_{x\to0}\frac{\sin^2(x)}{\tan(x)-x}=L. But \lim_{x\to0}\frac{\sin^2(x)}{\tan(x)-x}=\lim_{x\to0}\left\{\frac{1-\cos(x)}{\tan(x)-x}\cdot\left(1+\cos(x)\right)\right\}. And since \lim_{x\to0}\left\{1+\cos(x)\right\}=2, we may by assumption conclude that \frac{L}{2}=\lim_{x\to0}\frac{1}{1+\cos(x)}\cdot\l  im_{x\to0}\left\{\frac{1-\cos(x)}{\tan(x)-x}\cdot\left(1+\cos(x)\right)\right\} =\lim_{x\to0}\left\{\frac{1-\cos(x)}{\tan(x)-x}\cdot\frac{1+\cos(x)}{1+\cos(x)}\right\}=\lim_{x  \to0}\frac{1-\cos(x)}{\tan(x)-x} which tells us that \lim_{x\to0}\frac{1-\cos(x)}{\tan(x)-x} exists, but this is ludicrous since 1-\cos(x)\sim \frac{x^2}{2} and \tan(x)-x\sim\frac{-x^3}{3} . Thus, our assumption was false and the original limit does not exist.
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