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Math Help - Limit

  1. #1
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    Limit

    Evaluate

    <br /> <br />
\lim_{x\to0} \frac {(e^x)^2-1-x^2-\frac {x^4}{2}}{cosx-1+\frac {x^2}{2}-\frac {x^4}{4!}}<br />

    Thanks.
    Last edited by khatz; August 3rd 2009 at 06:12 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by khatz View Post
    Evaluate

    <br /> <br />
\lim_{x\to0} \frac {e^x-1-x^2-\frac {x^4}{2}}{cosx-1+\frac {x^2}{2}-\frac {x^4}{4!}}<br />

    Thanks.
    When you take this limit the way it is, you end up with the indeterminate case \frac{0}{0}

    So you can now apply L'H˘pital's rule.

    Thus, \lim_{x\to0} \frac {e^x-1-x^2-\frac {x^4}{2}}{cosx-1+\frac {x^2}{2}-\frac {x^4}{4!}}=\lim_{x\to0} \frac {e^x-2x-2x^3}{-\sin x-x-\frac {x^3}{3!}}

    However, when we evaluate the limit now, it approaches \frac{1}{0}, which is undefined.

    Does this help?
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  3. #3
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    Take taylor expand:
    e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+...+\frac{x^  {n}}{n!}+o(x^{n})<br />
cos(x)=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-...+(-1)^{n}\frac{x^{2n}}{(2n)!}+o(x^{2n+1})
    \lim_{x\to0} \frac {e^x-1-x^2-\frac {x^4}{2}}{cosx-1+\frac {x^2}{2}-\frac {x^4}{4!}}<br />
=\lim_{x\rightarrow 0}\frac{\frac{x^{3}}{3!}+...+\frac{x^{n}}{n!}}{-\frac{x^{6}}{6!}+...+(-1)^{n}\frac{x^{2n}}{(2n)!}}=\lim_{x\to0}-\frac{6!}{3!\cdot x^{3}}=\infty
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  4. #4
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    Quote Originally Posted by centry57 View Post
    Take taylor expand:
    e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+...+\frac{x^  {n}}{n!}+o(x^{n})<br />
cos(x)=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-...+(-1)^{n}\frac{x^{2n}}{(2n)!}+o(x^{2n+1})
    \lim_{x\to0} \frac {e^x-1-x^2-\frac {x^4}{2}}{cosx-1+\frac {x^2}{2}-\frac {x^4}{4!}}<br />
=\lim_{x\rightarrow 0}\frac{\frac{x^{3}}{3!}+...+\frac{x^{n}}{n!}}{-\frac{x^{6}}{6!}+...+(-1)^{n}\frac{x^{2n}}{(2n)!}}=\lim_{x\to0}-\frac{6!}{3!\cdot x^{3}}=\infty
    Sorry, I made a typo. The first term of the numerator should be (e^x)^2instead of e^x.
    And then the answer goes
    ... =\lim_{x\rightarrow 0}\frac{\frac{x^{6}}{3!}+\frac{x^{8}}{4!}+...+\fra  c{x^{2n}}{n!}}{-\frac{x^{6}}{6!}+...+(-1)^{n}\frac{x^{2n}}{(2n)!}}

    I wonder how you got your second last step?
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  5. #5
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    e^{2x}=1+\frac{2x}{1!}+\frac{(2x)^{2}}{2!}+...+\fr  ac{(2x)^{n}}{n!}+o((2x)^{n})
    cos(x)=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-...+(-1)^{n}\frac{x^{2n}}{(2n)!}+o(x^{2n+1})
    \lim_{x\to0} \frac {e^{2x}-1-x^2-\frac {x^4}{2}}{cosx-1+\frac {x^2}{2}-\frac {x^4}{4!}}
    =\lim_{x\to0}\frac{1+\frac{2x}{1!}+\frac{(2x)^{2}}  {2!}+...+\frac{(2x)^{n}}{n!}-1-x^2-\frac {x^4}{2}}{1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-...+(-1)^{n}\frac{x^{2n}}{(2n)!}-1-x^2-\frac {x^4}{2}}
    =\lim_{x\to0}\frac{1}{x}\cdot \frac{2+x+\frac{3x^{2}}{4}+\frac{x^{3}}{6}+...+\fr  ac{2^{n}(x)^{n-1}}{n!}}{-\frac{3}{2}-\frac{x^{2}}{4}+...+(-1)^{n}\frac{x^{2n-2}}{(2n)!}}
    =\lim_{x\to0}\frac{1}{x}\cdot \frac{2}{-\frac{3}{2}}=\infty
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  6. #6
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    Anyone can tell me how to edit a formular quickly?
    I only use this web :Online LaTeX Equation Editor
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