lim((x^2+4x+3)/(x+1)-(αx-β))=3,x->infinity

$\displaystyle{\lim_{x \to \infty}}\dfrac{x^2+4x+3}{x+1}-(\alpha x + \beta)=3$

$\displaystyle{\lim_{x\to \infty}}\dfrac{(x+3)(x+1)}{x+1}-(\alpha x+\beta)$

$\displaystyle{\lim_{x\to \infty}}(x+3)-(\alpha x+\beta)=3$

$\displaystyle{\lim_{x\to \infty}}(1-\alpha)x+(3-\beta)=3\\ \\ \alpha = 1 \\ \beta = 0$