Find the limit of the function Let we have the following function
$$F(x)=\frac{x^x-x}{\ln(x)-x+1}$$
Find $$\lim_{x \to 1}\frac{x^x-x}{\ln(x)-x+1}$$
 A: Hint: given the form of the limit, you can use L'Hopital's rule twice. 
A: You can use l'Hopital's rule since the limit of the top and bottom is 0. Call the numerator f(x) and the denominator g(x). Find f'(x) and g'(x) then find lim f'(x)/g'(x). If the limit still results in an indeterminate form you can repeat the process. Lim f(x)/g(x)= lim f'(x)/g'(x)
A: In a neighbourhood of the origin,
$$ (1+z)^{(1+z)}=\exp\left((1+z)\log(1+z)\right)=\exp\left((1+z)(z-\frac{z^2}{2}+O(z^3))\right)\\=\exp\left(z+\frac{z^2}{2}+O(z^3)\right)=1+z+z^2+O(z^3)$$
so the given limit equals:
$$\lim_{z\to 0}\frac{(1+z+z^2)-(1+z)}{\left(z-\frac{z^2}{2}\right)-z}=\color{red}{-2}.$$
A: $$\large L=\lim_{x \to 1}\frac{x^x-x}{\ln(x)-x+1}=\lim_{x \to 1}\frac{x^x(1+\log x)-1}{1/x-1}=\lim_{x \to 1}\frac{x^{x+1}+x^{x+1}\log x-x}{1-x}=\lim_{x \to 1}\frac{x^{x+1}+x^{x+1}\log x-x}{1-x}\\\large =\lim_{x \to 1}-(x^x(1+x+x\ln x)+x^x(1+\ln x)(1+x\ln x)-1)=-2$$

Notes:


*

*$$y=x^x\implies \ln y=x\ln x\implies y'/y=1+\ln x\implies y'=y(1+\ln x)$$

*$$y=x^{x+1}=x^x.x\implies y'=x^x(1+\ln x).x+x^x=x^x(1+x+x\ln x)$$

*$$y=x^{x+1}\ln x\implies y'=x^x(1+x+x\ln x).\ln x+x^{x+1}/x=x^x(1+\ln x)(1+x\ln x)$$

A: Letting $u=x^x$ we have $$\log u(x)=x\log x$$
$$ \frac 1 {u(x)} u'(x)=1+\log x$$
$$u'(x)={u(x)}(1+\log x)$$
Now, By L'Hospital Rule,
$$\lim_{x\to 1}\frac{x^x-x}{\log x-x-1}=\lim_{x\to 1}\frac{u(x)(1+\log x)-1}{\frac 1 x -1}=\lim_{x\to 1}\frac{u'(x)(1+\log x)+u(x)(1+\frac 1 x)}{\frac {-1}{x^2}}=\frac{u'(1)+2}{-1}=\frac {1+2}{-1}=-3$$
