[tex]f(x) = \dfrac{2^x-x^2}{x-2}\\ \\\\ \underline{\text{Asimptote verticale:}} \\ \\ \lim\limits_{x\to 2}\dfrac{2^x-x^2}{x-2} = \lim\limits_{x\to 2}\dfrac{2^x\ln 2 - 2x}{1}= 4\ln2 -4\neq \pm \infty \\ \\ \Rightarrow \text{Nu exista asimptote verticale.}\\ \\\\ \underline{\text{Asimptote orizontale:}}\\ \\ \lim\limits_{x\to \pm \infty}\dfrac{2^x-x^2}{x-2} = \lim\limits_{x\to \pm \infty}(2^x\ln 2 - 2x) = +\infty \\ \\ \Rightarrow \text{Nu exista asimptote orizontale.}[/tex]
[tex]\underline{\text{Asimptote oblice:}} \\ \\ \lim\limits_{x\to \pm \infty} \Big(\dfrac{2^x-x^2}{x-2}-mx-n\Big) =\\ \\ =\lim\limits_{x\to \pm \infty} \Big(\dfrac{2^x-x^2-mx^2+2mx}{x-2}\Big)-n =\\ \\ = \lim\limits_{x\to \pm \infty} \Big(\dfrac{2^x\ln 2-2x-2mx+2m}{1}\Big)-n\\ \\\\ \text{Observam ca daca x} \to \infty,\quad \text{limita intotdeauna va fi }+\infty[/tex]
[tex]\text{Deci nu avem asimptota oblica spre }+\infty.[/tex]
[tex]\Rightarrow \lim\limits_{x\to -\infty}\Big(2^x\ln 2-2x(1+m)+2m}\Big)-n = \\ \\ =0+\lim\limits_{x\to -\infty}\Big(-2x(1+m)+2m}\Big)-n = \\\\ \Rightarrow 1+m = 0 \text{ deoarece vrem ca limita sa fie finita} \Rightarrow m = -1 \\ \\= \lim\limits_{x\to -\infty} \Big(2\cdot(-1)\Big) - n = -2-n\\ \\ \Rightarrow -2-n = 0 \Rightarrow n = -2 \\ \\ \Rightarrow y = mx+n \Rightarrow \boxed{y = -x-2\,\,\big|\,\,\text{asimptota oblica spre }-\infty}[/tex]
