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Math Help - l'Hospital's rule and limits.

  1. #1
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    l'Hospital's rule and limits.

    \lim_{x \to \0}\frac{(3+x)^{\frac{4}{7}} - (3-x)^{\frac{4}{7}}}{x}, which gives the indeterminate form \frac{0}{0}. Using l'Hospital' rule gives:
    \lim_{x \to \0}[\frac{4}{7}(3+x)^{-\frac{3}{7}} - \frac{4}{7}(3+x)^{-\frac{3}{7}}] = 0

    But, the book gives a different answer and I confirmed their answer by using a different method. What I want to know exactly is why l'Hospital's rule failed to work for this limit.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: l'Hospital's rule and limits.

    Quote Originally Posted by Pupil View Post
    \lim_{x \to \0}\frac{(3+x)^{\frac{4}{7}} - (3-x)^{\frac{4}{7}}}{x}, which gives the indeterminate form \frac{0}{0}. Using l'Hospital' rule gives:
    \lim_{x \to \0}[\frac{4}{7}(3+x)^{-\frac{3}{7}} - \frac{4}{7}(3+x)^{-\frac{3}{7}}] = 0

    But, the book gives a different answer and I confirmed their answer by using a different method. What I want to know exactly is why l'Hospital's rule failed to work for this limit.
    The derivative of (3+x)^{\frac{4}{7}} - (3-x)^{\frac{4}{7}} is \frac{4}{7}(3+x)^{-\frac{3}{7}}+\frac{4}{7}(3-x)^{-\frac{3}{7}}.
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