I assume your limit is
[tex]\displaystyle \lim_{x\to4} (3x-3)^{5/2}[/tex]
Recall that [tex]\sqrt x = x^{1/2}[/tex] 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 :
[tex]\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}[/tex]
The Limit of (3 x minus 3) Superscript five-halves Baseline as x approaches 4 is 243.
A limit is a value that the given function approaches as the input value approaches to some value. Limits are essential to calculus and mathematical analysis and are used to define continuity, derivatives, and integrals.
We have;
[tex]\lim_{x \to 4} (3x-3)^{5/2} \\ \lim_{x \to 4} 3(x-1)^{5/2}\\3 \lim_{x \to 4} (x-1)^{5/2}\\3(3)^{5/2}\\=243.[/tex]
Learn more about limits:
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