[tex]\text{Aplicand teorema lui l'Hopital avem:}\\
\displaystyle\limits\lim_{x\to 2}\dfrac{x^x-2^x}{2^x-4}=\displaystyle\limits\lim_{x\to 2}\dfrac{x^x(\ln x+1)-2^x\cdot \ln2}{2^x\cdot \ln 2}=\displaystyle\limits\lim_{x\to 2}
\left(\dfrac{x}{2}\right)^x\cdot \dfrac{\ln x+1}{\ln 2}-1=\\ =\left(\dfrac{2}{2}\right)^2\cdot \dfrac{\ln 2+1}{\ln 2}-1=\dfrac{\ln 2+1-\ln 2}{\ln 2}=\boxed{\dfrac{1}{\ln 2}}[/tex]
