Answer
$\lim\limits_{t\to\infty}\frac{\sqrt t+t^2}{2t-t^2}=-1$
Work Step by Step
$$A=\lim\limits_{t\to\infty}\frac{\sqrt t+t^2}{2t-t^2}$$
Divide both numerator and denominator by $t^2$
$A=\lim\limits_{t\to\infty}\frac{\frac{\sqrt t}{t^2}+\frac{t^2}{t^2}}{\frac{2t}{t^2}-\frac{t^2}{t^2}}$
$A=\lim\limits_{t\to\infty}\frac{\sqrt\frac{t}{t^4}+1}{\frac{2}{t}-1}$
$A=\lim\limits_{t\to\infty}\frac{\sqrt\frac{1}{t^3}+1}{\frac{2}{t}-1}$
$A=\lim\limits_{t\to\infty}\frac{\frac{1}{t^{3/2}}+1}{\frac{2}{t}-1}$
$A=\frac{\lim\limits_{t\to\infty}(\frac{1}{t^{3/2}}+1)}{\lim\limits_{t\to\infty}(\frac{2}{t}-1)}$
$A=\frac{0+1}{2\times0-1}$
$A=-1$
