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Compute the limit of the given function

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 1^-} \big( (\log x)(\log (1-x)) \big). \]

To evaluate this limit we rewrite it in the indeterminate form -\infty / -\infty and apply L’Hopital’s rule,

    \begin{align*} \lim_{x \to 1^-} \big( (\log x)(\log (1-x)) \big) &= \lim_{x \to 1^-} \frac{\log (1-x)}{\frac{1}{\log x}} \\[9pt] &= \lim_{x \to 1^-} \frac{\frac{-1}{1-x}}{\frac{-1}{x (\log x)^2}} \\[9pt] &= \lim_{x \to 1^-} \frac{-x (\log x)^2}{1-x} \\[9pt] &= \lim_{x \to 1^-} - \frac{(\log x)^2 + 2 \log x}{-1} &(\text{L'Hopital again}) \\[9pt] &= \lim_{x \to 1^-} \big( (\log x)^2 + 2 \log x \big) \\[9pt] &= 0. \end{align*}

One comment

  1. Anonymous says:

    So, Apostol explicitly says that he’s not going to introduce forms lof L’Hopital’s rule for expressions of the form infinity/infinity. He says this on the bottom of page 300. So I don’t think it’s in keeping with the spirit of the text to use L’Hopital here.

    The solution that I think Apostol may be looking for here is by approximation with a Taylor polynomial. If you substitute x = t + 1, then log(1 – x) = log t, and you can get a nice Taylor polynomial for log x = log (t + 1). Then apply the result from Example 2 on page 302.

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