Salut,
Aplicăm regula lui L'Hospital:
[tex]L=\lim\limits_{x\to +\infty}\dfrac{2x}{ln(1+e^x)}=\lim\limits_{x\to +\infty}\dfrac{(2x)'}{[ln(1+e^x)]'}=\lim\limits_{x\to +\infty}\dfrac{2}{\dfrac{(1+e^x)'}{1+e^x}}=\\\\\\=\lim\limits_{x\to +\infty}\dfrac{2(1+e^x)}{e^x}=\lim\limits_{x\to +\infty}\left(\dfrac{2}{e^x}+2\right)=0+2=2,\ deci\ L=2.[/tex]
Ai înțeles rezolvarea ?
Green eyes.
