Find the value of the following limit.
![\[ \lim_{x \to 1^+} \frac{x^x - x}{1 - x + \log x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e214c871171ab2a6b86390d125c03f0c_l3.png "Rendered by QuickLaTeX.com")
We apply L’Hopital’s rule (twice) to compute the limit (Noting that, 
),
![\begin{align*} \lim_{x \to 1^+} \frac{x^x - x}{1 - x + \log x} &= \lim_{x \to 1^+}\frac{x^x + x^x \log x - 1}{\frac{1}{x} -1} \\[9pt] &= \lim_{x \to 1^+} \frac{x^{x+1} + x^{x+1} \log x - x}{1 - x} \\[9pt] &= \lim_{x \to 1^+} \frac{x^{x+1} \left( \log x + \frac{x+1}{x} \right) + x^{x+1} \left( \log x + \frac{x+1}{x} \right) + x^x - 1}{-1} \\[9pt] &= -2. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e97034be0adcc38c6932d48975490d97_l3.png "Rendered by QuickLaTeX.com")
Since this limit exists, the application of L’Hopital’s rule was justified.
Find the value of the following limit.
We apply L’Hopital’s rule (twice) to compute the limit (Noting that, ),
Since this limit exists, the application of L’Hopital’s rule was justified.