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Math Help - trigonometric limits

  1. #1
    Member purplec16's Avatar
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    trigonometric limits

    Evaluate the following limit:

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

    Ok i really don't know what to do

    so were trying to get \sin {x}^{2 }
    Last edited by purplec16; May 25th 2011 at 03:54 PM. Reason: error with latex
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  2. #2
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    Use the result

    \displaystyle \lim_{x\to 0 } \frac{\sin x}{x} = 1
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  3. #3
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    Quote Originally Posted by purplec16 View Post
    Evaluate the following limit:

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

    Ok i really don't know what to do

    so were trying to get \sin {x}^{2 }
    Another way to think about this is in terms of the derivative.

    Let

    f(x)=\frac{\sin(x)}{x}=\frac{\sin(x)-0}{x-0}

    so the limits is

    \lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)

    but each factor is just the definition of the derivative evaluates at zero

    but

    \frac{d}{dx}\sin(x) \bigg|_{x=0}=1

    \lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)=1\cdot 1=1
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  4. #4
    Member purplec16's Avatar
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    Quote Originally Posted by pickslides View Post
    Use the result

    \displaystyle \lim_{x\to 0 } \frac{\sin x}{x} = 1
    I know that but how do I make the numerator x^2
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  5. #5
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    \displaystyle  \frac{\sin^2 x}{x^2} =   \frac{\sin x}{x}\times   \frac{\sin x}{x}
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  6. #6
    Member purplec16's Avatar
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    wow... so true...i can't believe i didnt see that -.-

    thus the answer is 1
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  7. #7
    Member purplec16's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Another way to think about this is in terms of the derivative.

    Let

    f(x)=\frac{\sin(x)}{x}=\frac{\sin(x)-0}{x-0}

    so the limits is

    \lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)

    but each factor is just the definition of the derivative evaluates at zero

    but

    \frac{d}{dx}\sin(x) \bigg|_{x=0}=1

    \lim_{x \to 0}[f(x)]^2=\left( \lim_{x \to 0} f(x)\right)\left( \lim_{x \to 0} f(x)\right)=1\cdot 1=1

    Thats a really complex way to put it but i understand
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