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

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

 

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

 

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

 

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

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