f(x)=x+lnx, f:(0;∞)→R. Calculam limita la ∞ si ne da ∞+∞=∞, deci nu are asimtota orizontala la +∞. Cautam asimtota oblica , Panta m=[tex] \lim_{x \to \infty} \frac{f(x)}{ x}=1+ \lim_{x\to \infty} \frac{lnx}{x}= (\frac{\infty}{\infty})=l'H=1+lin \frac{1}{x} =1+0=1 [/tex]. Cautam panta n=[tex] \lim_{x \to \infty} (f(x)+mx)= \lim_{x \to \infty} (x+lnx-x)= \lim_{ \to \infty} lnx= [/tex]∞, deci nu avem asimtota oblica la ∞. Asimtota oblica exista si are ecuatia y=mx+n, numai daca m si n exista si sunt finite. Functia ta are asimtota verticala de ecuatie x=0.
