There is a Limit to Everything

Since the Limit is of the form $\dfrac{0}{0}$ , we can use L'Hopital's Rule.

$L = \displaystyle \lim_{x\to 1} \dfrac{x - x^x}{1-x+\ln x} = \lim_{x \to 1} \dfrac{1 - x^x(1+\ln x)}{-1 + \frac{1}{x}}$

This is again of the $\dfrac{0}{0}$ form, so L'Hopital's Rule can be applied again.

$L = \displaystyle \lim_{x \to 1}\dfrac{1 -x^x(1+ \ln x)}{-1 + \frac{1}{x}} = \lim_{x \to 1} \dfrac{-x^x(1+ \ln x)^2 - x^x\left(\frac{1}{x}\right) }{\frac{-1}{x^2}} = \dfrac{-1 + -1}{-1}$

$\Rightarrow L = \boxed{2}$

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Note:

• The derivative of $x^x$ with respect to $x$ is found in the following manner.

$\text{Let } y = x^x$

$\Rightarrow \ln y = x \ln x \Rightarrow \dfrac{1}{y} \dfrac{dy}{dx} = \dfrac{x}{x} + \ln x$

$\Rightarrow \dfrac{dy}{dx} = y(1 + \ln x) = x^x(1 + \ln x)$

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Without using L'Hôpital's rule (we can, of course, just use twice and the solution comes out), let $t=x-1$ , so we have