[tex]\displaystyle f:(-\infty,1)\rightarrow(1,\infty),f(x)=\frac{x^2+mx+1}{x-1}[/tex]
[tex]y=x+2\text{ asimptot\u a oblic\u a}\Rightarrow \lim\limits_{n\rightarrow\infty\displaystyle}\frac{f(x)}{x}=1 \text{ \c si }\lim\limits_{x\rightarrow\infty}\left[f(x)-x\right]=2\\\\\displaystyle\lim\limits_{x\rightarrow\infty}\frac{\frac{x^2+mx+1}{x-1}}{x}=\lim\limits_{x\rightarrow\infty}\frac{x^2+mx+1}{x^2-x}\overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\rightarrow\infty}\frac{x^2\left(1+\frac{m}{x}+\frac1{x^2}\right)}{x^2\left(1-\frac1x\right)}=1,\forall m\in\mathbb{R}[/tex]
[tex]\displaystyle\lim\limits_{x\rightarrow\infty}\left(\frac{x^2+mx+1}{x-1}-x\right)=\lim\limits_{x\rightarrow\infty}\frac{x^2+mx+1-x(x-1)}{x-1}=\lim\limits_{x\rightarrow\infty}\frac{x^2+mx+1-x^2+x}{x-1}=\lim\limit_{x\rightarrow\infty}\frac{(m+1)x+1}{x-1}\overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\rightarrow\infty}\frac{x\left(m+1+\frac1x\right)}{x\left(1-\frac1x\right)}=m+1=2\Rightarrow m=1[/tex]
