Question

What is Limit of (3 x minus 3) Superscript five-halves Baseline as x approaches 4? RootIndex 5 StartRoot 31 EndRoot RootIndex 5 StartRoot 92 EndRoot 15 243

Answer 1

I assume your limit is

$\displaystyle \lim_{x\to4} (3x-3)^{5/2}$

Recall that $\sqrt x = x^{1/2}$ is defined and continuous for all x ≥ 0, so the limand in this case is continuous for all 3x - 3 ≥ 0 or x ≥ 1. Then we can evaluate the limit by directly substituting x = 4 :

$\displaystyle \lim_{x\to4}(3x-3)^{5/2} = \left(3\lim_{x\to4}(x-1)\right)^{5/2} = \left(3^2\right)^{5/2} = 3^5 = \boxed{243}$