L’Hôpital’s rule, question 19

Two students evaluate $\displaystyle\lim_{x \to 1} \frac{x\log x - x + 1}{x^{2} - 1}$ as shown. Which solution is right, and why?

\begin{aligned} &\textbf{Solution A} \\ \text{A1:}\quad & \text{at } x = 1: \ \frac{0 - 1 + 1}{1 - 1} = \frac{0}{0} \\[4pt] \text{A2:}\quad & \overset{\text{l'H}}{=} \lim_{x \to 1} \frac{\log x + 1 - 1}{2x} = \lim_{x \to 1} \frac{\log x}{2x} \\[4pt] \text{A3:}\quad & = \frac{\log 1}{2} = \frac{0}{2} = 0 \\[14pt] &\textbf{Solution B} \\ \text{B1:}\quad & \text{at } x = 1: \ \frac{0}{0}, \text{ so l'Hôpital's rule applies} \\[4pt] \text{B2:}\quad & \overset{\text{l'H}}{=} \lim_{x \to 1} \frac{\log x}{2x} \\[4pt] \text{B3:}\quad & \overset{\text{l'H}}{=} \lim_{x \to 1} \frac{1/x}{2} \\[4pt] \text{B4:}\quad & = \frac{1}{2} \end{aligned}

A is right; B uses l’Hôpital’s rule a second time on log x/(2x), which tends to 0/2, a form that is not indeterminate.
B is right; A stops while the top still tends to 0, and a zero on top means the limit is not settled yet.
Both are right: each differentiates the top and the bottom correctly, and l’Hôpital’s rule may be applied as often as you like.
Neither is right: l’Hôpital’s rule needs the quotient rule on the whole fraction, and both differentiate the top and bottom separately.

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