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Math Help - Finding Limit using L'Hopital's Rule

  1. #1
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    Finding Limit using L'Hopital's Rule

    How do you find these limits using L'Hopital's rule?

    1.  \lim_{x \to 0}~\frac{sin x} {x^{1/3}}

    2. \lim_{x \to 0}~\frac{x} {(sin x)^{1/3}}

    In both cases, when I apply L'Hopitals rule, the derivative of the denominator won't allow me to solve because there ends up being a zero in there. ie 1/3x^{-2/3}

    Thanks.
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  2. #2
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    In both cases, you should have a negative power in the denominator which moves it up to the numerator. Be sure that you simplify the expression before evaluating.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by suboba View Post
    How do you find these limits using L'Hopital's rule?

    1.  \lim_{x \to 0}~\frac{sin x} {x^{1/3}}

    2. \lim_{x \to 0}~\frac{x} {(sin x)^{1/3}}

    In both cases, when I apply L'Hopitals rule, the derivative of the denominator won't allow me to solve because there ends up being a zero in there. ie 1/3x^{-2/3}

    Thanks.
    You can actually do these without L'Hopitals Rule (but you need to know that \lim_{x\to0}\frac{\sin x}{x}=1 ):

    1. \lim_{x\to0}\frac{\sin x}{x^{\frac{1}{3}}}=\lim_{x\to0}x^{\frac{2}{3}}\cd  ot\frac{\sin x}{x}=\lim_{x\to 0}x^{\frac23}\cdot\lim_{x\to0}\frac{\sin x}{x}=0\cdot1=\boxed{0}

    2. \lim_{x\to0}\frac{x}{\left(\sin x\right)^{\frac{1}{3}}}=\lim_{x\to0}\left(\sin x\right)^{\frac23}\cdot\frac{x}{\sin x}=\lim_{x\to0}\left(\sin x\right)^{\frac{2}{3}}\cdot\lim_{x\to0}\frac{x}{\s  in x} =\lim_{x\to0}\left(\sin x\right)^{\frac{2}{3}}\cdot\frac{1}{\lim\limits_{x  \to0}\displaystyle\frac{\sin x}{x}}=0\cdot\frac{1}{1}=\boxed{0}
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