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Math Help - Limits question L' Hopitals Rule

  1. #1
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    Limits question L' Hopitals Rule

    lim x--> 0

    (1-cosx)^{3/2}/(x-sinx)


    I did LHop rule like 2 or 3 times and then split it into two terms, I got that both terms go to zero
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  2. #2
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    Re: Limits question L' Hopitals Rule

    \begin{align*}\lim_{x \to 0} \dfrac{(1-\cos x)^{3/2}}{x-\sin x} & = \lim_{x \to 0} \dfrac{\tfrac{3}{2}\left(1-\cos x\right)^{1/2}\sin x}{1-\cos x} \\ & = \dfrac{3}{2}\lim_{x \to 0} \dfrac{\sin x}{(1-\cos x)^{1/2}}\cdot \dfrac{(1+\cos x)^{1/2}}{(1+\cos x)^{1/2}} \\ & = \dfrac{3}{2} \lim_{x \to 0} \dfrac{\sin x(1+\cos x)^{1/2}}{(1-\cos^2 x)^{1/2}} \\ & = \dfrac{3}{2}\lim_{x \to 0} \dfrac{\sin x(1+\cos x)^{1/2}}{\sin x} \\ & = \dfrac{3}{2}\lim_{x \to 0} (1+\cos x)^{1/2} \\ & = \dfrac{3\sqrt{2}}{2}\end{align*}

    I only used L'Hospital's rule once (the first step).
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