\(\displaystyle\lim_{t\to\infty}\dfrac{\sqrt t+t^2}{4t-t^2}=\lim_{t\to\infty}\dfrac{\dfrac1{t^{3/2}}+1}{\dfrac4t-1}=\lim_{t\to\infty}\dfrac1{-1}=-1\)
Or, if you actually did mean to find the limit as \(x\to\infty\), then the result would simply be the limand \(\dfrac{\sqrt t+t^2}{4t-t^2}\), since this expression doesn't depend on \(x\) (as far as can be seen).
