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Math Help - Limit

  1. #1
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    Limit

    How do you solve (1 - (cosx)^3)/x^2 as x -> 0? I tried to use l'Hospital rule but it didn't help.
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  2. #2
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    no , l'Hospital rule should help you.

    use it two times and you can get the answer
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  3. #3
    MHF Contributor Calculus26's Avatar
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    you only need L'Hopitals Rule once:

    (1 - (cosx)^3)/x^2

    differentiate:

    3cos^2(x) {sin(x)/2x}

    and


    lim 3cos^2(x) {sin(x)/2x} = 3/2 as lim sin(x)/x = 1
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by antero View Post
    How do you solve (1 - (cosx)^3)/x^2 as x -> 0? I tried to use l'Hospital rule but it didn't help.
    You can calculate this limit without L'Hospital's rule:

    \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}x}}<br />
{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \left( {1 - {{\sin }^2}x} \right)\cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x + {{\sin }^2}x\cos x}}{{{x^2}}} =

    = 2\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}<br />
{2x^2} + \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}<br />
{{{x^2}}}\cos x = 2\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}} + {\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}} \right)^2}\mathop {\lim }\limits_{x \to 0} \cos x =

    = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{\frac{{{x^2}}}{4}}} + 1 = \frac{1}{2}{\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2} + 1 = \frac{1}{2} + 1 = \frac{3}{2}.
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