Răspuns:
m=-3
Explicație pas cu pas:
[tex]f(x)-x =\frac{x^2+3x+m}{x-m} -x \\= \frac{x^2+3x+m-x(x-m)}{x-m}\\= \frac{x^2+3x+m-x^2+mx}{x-m}\\\\= \frac{3x+m+mx}{x-m} \\ => f(x)\\ \lim_{x \to \infty} f(x)= \lim_{x\to\infty} \frac{3x+mx+m}{x-m} (exceptie \frac{\infty}{\infty} ) L'Hopital => \lim_{x \to \infty} \frac{(3x+mx+m)'}{(x-m)'}= \lim_{x \to \infty} \frac{3+m}{1}.\\stiim ca \lim_{x \to \infty} (f(x)-x) =0 => 3+m=0 => m=-3[/tex]
