Question 980460
t^3 - 27 factors to (t-3)(t^2 + 3t + 9) when you use the difference of cubes factoring rule


So {{{(t^3-27)/(t-3)=((t-3)(t^2 + 3t + 9))/(t-3)=t^2 + 3t + 9}}}


Now plug in t = 3 {{{t^2 + 3t + 9=3^2+3(3)+9 = 27}}}


So the limiting value is 27