aufgabe 1a

$$\begin{array}{rl}\underset{x\to 1}{\lim }\frac{x^x-x}{1-x+\ln x}& \stackrel{\text{L’H}}{\underset{0/0}{=}}\underset{x\to 1}{\lim }\frac{x^x\cdot (\ln x+1)-1}{0-1+\frac{1}{x}}\\ & \stackrel{\text{L’H}}{\underset{0/0}{=}}\underset{x\to 1}{\lim }\frac{x^x\cdot (\ln x+1{)}^2+x^x\cdot \frac{1}{x}}{-\frac{1}{x^2}}\\ & =-2\end{array}$$

Ableitung von x hoch x

$$\begin{array}{rl}(x^x{)}^{\prime }& ={\left(e^{x\ln x}\right)}^{\prime }\\ & =e^{x\ln x}\cdot (x\ln x{)}^{\prime }\\ & =e^{x\ln x}\cdot \left((x{)}^{\prime }\cdot \ln x+x\cdot (\ln x{)}^{\prime }\right)\\ & =e^{x\ln x}\cdot \left(1\cdot \ln x+x\cdot \frac{1}{x}\right)\\ & =e^{x\ln x}\cdot \left(\ln x+1\right)\\ & =x^x\cdot \left(\ln x+1\right)\end{array}$$

Ableitung von 1 durch x

$$\begin{array}{rl}{\left(\frac{1}{x}\right)}^{\prime }& ={\left(x^{-1}\right)}^{\prime }\\ & =-1\cdot x^{-2}\\ & =-\frac{1}{x^2}\end{array}$$