How find this $\lim_{x\to 1} (\frac{x^x-x}{\ln{x}-x+1}+2)/( x-1)$ Find  the limit
$$I=\lim_{x\to 1}\dfrac{\left(\dfrac{x^x-x}{\ln{x}-x+1}+2\right)}{x-1}$$
I want let $x-1=t$
$$I=\lim_{t\to 0}\dfrac{\dfrac{(t+1)^{t+1}-(t+1)}{\ln{(t+1)}-t}+2}{t}=\lim_{t\to  0}\dfrac{(t+1)^{t+1}-(t+1)+2\ln{(t+1)}-2t}{t(\ln{(t+1)}-t)}=\lim_{t\to 0}\dfrac{(t+1)^{t+1}+2\ln{(t+1)}-3t-1}{-\dfrac{t^3}{2}}$$
I have  solve this result $-\dfrac{7}{3}$by hand ,is true ?

Taylor series, built at $x=1$ could work $$\log(x)=(x-1)-\frac{1}{2} (x-1)^2+\frac{1}{3} (x-1)^3+O\left((x-1)^4\right)$$ $$\log(x)-x+1=-\frac{1}{2} (x-1)^2+\frac{1}{3} (x-1)^3+O\left((x-1)^4\right)$$ $$x^x-x=(x-1)^2+\frac{1}{2} (x-1)^3+O\left((x-1)^4\right)$$ Combining the pieces $$\frac{\left(\frac{x^x-x}{\ln{x}-x+1}+2\right)}{x-1}=-\frac{7}{3}-\frac{11 }{9}(x-1)-\frac{83}{135} (x-1)^2+O\left((x-1)^3\right)$$
