x=lne^x Limita devine
x→+∞ L=lim[ln(1+x²)-lne^x]=lim[ln(x²+1)/e^x]=ln lim (1+x²)/e^x]=ln∞/∞.Aplici teorema lui L`Hpspital
L=ln[lim(1+x²)/e^x]=ln lim[(1+x²) `/(e^x) `]=ln2x/e^x=ln∞/∞
Aplici in continuare T H`Lospital
L=lnlim(2x)`/(e^x)`=ln lim2/e^x=ln0=-∞
