Question

how you find the limit of (√t + t²)/4t-t² as x approaches positive infinity?

Answer 1

$\displaystyle\lim_{t\to\infty}\dfrac{\sqrt t+t^2}{4t-t^2}=\lim_{t\to\infty}\dfrac{\dfrac1{t^{3/2}}+1}{\dfrac4t-1}=\lim_{t\to\infty}\dfrac1{-1}=-1$

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).