The values of $\alpha \text{ and }\beta$ such that

$\underset{x\to \mathrm{\infty }}{lim} \left[\frac{x^2+1}{x-1}-\alpha x-2\beta \right]=\frac{3}{2}\text{ are }$

$\underset{x\to \mathrm{\infty }}{lim} \left[\frac{x^2+1}{x-1}-\alpha x-2\beta \right]$

$=\underset{x\to \mathrm{\infty }}{lim} \frac{x^2(1-\alpha )-(2\beta -\alpha )x+1+2\beta }{x-1}$

Since the last exists so $1-\alpha =0\Rightarrow \alpha =1$. In this
case the last limit is equal to

$\underset{x\to \mathrm{\infty }}{lim} \frac{-(2\beta -\alpha )+\frac{1+2\beta }{x}}{1-\frac{1}{x}}=-(2\beta -\alpha )=\frac{3}{2}$

$\Rightarrow -2\beta =\frac{3}{2}-1=\frac{1}{2}\Rightarrow \beta =-\frac{1}{4}$.