Divide through the limand by the highest power of t :
[tex]\displaystyle \lim_{t\to\infty} \frac{t^3-t+2}{(2t-1)(t^2+t+1)} = \lim_{t\to\infty} \frac{1 - \frac1{t^2} + \frac2{t^3}}{\left(2-\frac1t\right) \left(1+\frac1t + \frac1{t^2}\right)}[/tex]
(I distributed 1/t³ in the denominator as 1/t (2t - 1) and 1/t² (t² + t + 1).)
As to goes to infinity, these 1/tⁿ terms will converge to 0, and you're left with
[tex]\displaystyle \lim_{t\to\infty} \frac{1 - \frac1{t^2} + \frac2{t^3}}{\left(2-\frac1t\right) \left(1+\frac1t + \frac1{t^2}\right)} = \frac{1 - 0 + 0}{(2-0)(1+0+0)} = \boxed{\frac12}[/tex]
