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Math Help - Limits with trigonometric functions.

  1. #1
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    Limits with trigonometric functions.

    I am having trouble with this limit. I don't need an answer, but I would appreciate if someone could give me a hint or show me what I am missing.

    The following is what I can make sense of it so far:

    \lim_{x \to 0}\frac{tan\theta - sin\theta}{sin^3\theta}


    \rightarrow \lim_{x \to 0}\frac{\frac{sin\theta}{cos\theta} - sin\theta}{sin^3\theta}


    \rightarrow \lim_{x \to 0}\left(\frac{sin\theta}{cos\theta}\right)\left( \frac{1}{sin^3\theta}\right) - \frac{sin\theta}{sin^3\theta}


    \rightarrow \lim_{x \to 0}\frac{1}{cos\theta cos^2\theta} - \frac{1}{sin^2\theta}


    \rightarrow \lim_{x \to 0}\frac{1- cos\theta}{cos\theta sin^2\theta}


    I am aware that \lim_{x \to 0}\frac{1- cos x}{x} = 0; however, I am a little lost on how to get there, or if that is even the direction I should be taking. Any hints or explanations would be greatly appreciated. Thanks!

    Take care.
    /alan
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  2. #2
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    Re: Limits with trigonometric functions.

    Factor out a \sin(\theta) from both the numerator and denominator of your original ratio, and see where you can go from there.
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    Re: Limits with trigonometric functions.

    I believe you will need to apply L'H˘pital's rule.
    	\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
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  4. #4
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    Re: Limits with trigonometric functions.

    \frac{\tan{x} - \sin{x}}{\sin^3{x}} =

    \frac{\sin{x} - \cos{x}\sin{x}}{\cos{x}\sin^3{x}} =

    \frac{1 - \cos{x}}{\cos{x}\sin^2{x}} =

    \frac{1 - \cos{x}}{\cos{x}(1-\cos^2{x})} =

    \frac{1}{\cos{x}(1+\cos{x})}
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  5. #5
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    Re: Limits with trigonometric functions.

    Quote Originally Posted by AltF4 View Post
    I believe you will need to apply L'H˘pital's rule.
    	\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
    why don't you show the OP that method?
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