Question

Mă puteți ajuta la acest exercițiu

Answer 1

$\displaystyle f(1)=2\Rightarrow\frac{a+b+c}{-1}=2\Rightarrow a+b+c=-2\\\displaystyle\lim_{x\rightarrow\infty}\frac{f(x)}{x}=\lim_{x\rightarrow\infty}\frac{ax^2+bx+c}{x(x-2)}=\lim_{x\rightarrow\infty}\frac{ax^2+bx+c}{x^2-2x}\overset{\frac{\infty}{\infty}}{=}\lim_{x\rightarrow\infty}\frac{x^2\left(a+\frac{b}{x}+\frac{c}{x^2}\right)}{x^2(1-\frac{2}{x})}=a\Rightarrow a=2$

$\displaystyle\lim_{x\rightarrow\infty}\left[f(x)-2x\right]=\lim_{x\rightarrow\infty}\frac{2x^2+bx+c-2x(x-2)}{x-2}=\lim_{x\rightarrow\infty}\frac{(4+b)x+c}{x-2}\overset{\frac\infty\infty}{=}4+b=3\Rightarrow b=-1\\2-1+c=-2\Rightarrow c=-3\\\displaystyle f(x)=\frac{2x^2-x-3}{x-2}$